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-I £ L^J_
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""^^ILE tZl"t^ ^u^
THE
CARPENTER'S NEW GUIDE;
A
COMPLETE BOOK OF LINES
FOR
CARPENTRY AND JOINERY i ^
TREATING FULLY ON
SOFFITS, GROINS, NICHES, ROOFS, AND DOMES,
AND CONTAINING
A GREAT VARIETY OP ORIGmAL DESIGNS.
ALSO A FULL EXEMPLIFICATION OF
THE THEORY AND PRACTICE OP STAIR BUILDING, >
COKNICES, MOULDINGS, AND DRESSINGS OF EVERY DESORS?{jo|f.^^^v
INCLUDING ALSO » * l\ ^.^ ^
_ l\ TP -O -^ -^"^ 1
8(.)NUS. OBSERVATIONS AM) CALCULATIONS ON THE STRENGTH; Cff^»m^R.
BY PETER NICHOLSON, '' '. '. ^'''
AUTHOB OF "the carpbmtbr's ahd joiheb'b asbistakt/' ^'tub student's instbuctor to the pite obdebSi" etc.
r '
• «
BY N. K. DAVIS.
AND CONTAINING NUMEROUS NEW, IMPROVED, AND
ORIGINAL DESIGNS FOR ROOFS, DOMES, ETC.,
/
TUB WHOLE BEINO CAREFULLY AND THOROUOHLY REVISED /
BY SAMUEL SLOAN, ARCHITECT,
AurnoR OF "tof. mopkl AUcniTBcir/*
-^-«*»*>-
SIXTEENTH EDITION.
-<«-«-»»■•-
• • • - • • ••• ••
• • • • •
••*• ••• .*. •••*• •
!!••! • • •• • • •
..::.•:•• •• • •
PHILADELPHIA:
J. B. LIPPINCOTT & COMPANY.
18 6 7.
I . kJ
: I!lV/ YCr.K
I PULi^'iC LIBRARY
278087xV
ASTX>R, LEN#X ANB
LTli-ii-N ftUNDATlOMS
EDtcred; according to Act of Congress, in the year 1853, by
LIPPINCOTT, GRAMBO & CO.,
in the Clerk's Office of the District Court of the United States for the Eastern District of Pennsylvania
vnntionPKD bt j. paoait.
• • •"
•• ••
THE AUTHOR'S PREFACE.
To ft lK)ok intended merely for the use of Practical Meclianica, much Preface is not neces-
sary. It is proper, however, to any, that whatever rules by previous authors have on
examination proved to be true and well explained, these have been selected and adopted,
with such alterations aa a very close attention has warranted for the more easily compre-
hending them, for their greater accuracy or facility of application ; added to these, are many
examples which are entirely of my own inventionj and such as will, I am persuaded, conduce
very much to the accuracy of the work, and to the ease of the workman.
The arrangement of the subjects in this work is gradual and regular, and such aa afitudent
should pursue who wishes to attain a thorough knowledge of his profession : and as it if
Geometry that lays down all the first principles of building, measures of lines, angles, rind
Kolids, and gives rules for describing the various kinds of figures used in buildings; therefore,
as a necessary introduction to the art treated of, I have first laid down, and explained in the
terms of workmen, such problems of Geometry as are absolutely requisite to the well under-
standing and putting in practice the necessary lines for Carpentry. These problems duly
considered, and their results well understood, the learner may proceed to the theoretical part
of the subject, iu which SofBts claim particular attention ; for, by a thorough knowledge of
these, the student will be enabled to lay down arches which shall stand exactly perpen-
dicular over their plan, whatever form the plan may be: on this depends the well executing
all groins, arches, niches, &c., constructed in circular walla, or which stand up<»n irregular
bases; wherefore the importance of rightly understanding these I cannot too much insist
upon, their construction being so various and intricate, and their uses so frequently required.
The two plates of cuneoidal or winding soffits are new, and are constructed iu a more simple
and more accurate manner: yet this method is only a nearer approximation to truth than
the former one; the surface of a conchoid cannot be developed; that is, it cannot be ex-
tended on a plane : it is therefore absurd to look for perfection on this subject.
The next subject which regularly presents itself is Groins; for the construction of which
there will Ije found many methods entirely new; and besides the common figures, I have
bhown many which are difficult of execution, and not to be found in any other author. 1
»ve displayed a variety of methods for constructing -spherical niches, a form more frequenti}
d than the elliptic, which only has yet been explained.
(iii;
IT THE AUTHOR S PREFACE.
Among the varioa« methods for finding the Lines for Eoofe, I have given an entire new
one for finding the down and sride beveb of purlines. so that they shall exactlv fit against
the hip rafter; and by the same method the Jack rafter will be made to fit.
Of Dromes and Polygons, I have shown an entirely new method for finding their covering,
within the space of the board, thereby avoiding the tedious and ihcommodious method of
finding the lines on the dome itself^ as has been ahvays practised heretofore : also a method
for finding the form of the boards near the bottom, wlien a dome is to be covered horizon-
tally. Of dome-lights over staircases, or in the centre of groins, a rule upon true principles
i-* given, for finding their proper cur\'e against the wall, and the curve of the ribs; this has
never before Ijeen in.'ule public.
Hav::?g gr^ne thiis far in the Art of Carpentiy, it is necessary for me to say, by way of
caution and guard l/> the anient theori.-t, that there are s^jme surfaces which cannot be deve-
lofxrd; such as spherical or superoidical domes, where their coverings cannot be found by
any other fneans than by supposing the curved surface to become polygonal ; in which casi*
such domes may lie covered upfm true principles, as may be demonstrated. Let ns suppose*
a polygonal dome inscribed in a spherical one; then, the greater the number of sides of the
polygonal dome, the nearer it will coincide with its circumscribing spherical one. Again,
let us suppose that this polygonal dome has an infinite number of sides; then, its surface
will exactly coincide with the spherical dome, and therefore in anything which we shall
have ocr^asion to practise, this method will be sufficiently near; as, for example, in a dome
of one hundre^l sides, of a for^t each, the rule for finding such a covering will give the practice
s^> very near, that the variation from absolute truth could not be perceived.
Having gone through the constructive part of Carpentry, I next proceed to examples
showing the Ix^st forms of floors, partitions, trusses for roofs, truss girders, domes, &c., which
shall resist their own weight, or the addition of any adventitious load.
To cfjnclude : as I pretend not to infallibility, I hope to be judged with candor, being
always o[jen to conviction, from a knowledge of the difficulty and intricacy of science; yet
I liO|H* that my labcjrs niay be of s<jm<5 use to others in shortening the road, and smoothing
the path through which, for many years, I have been a persevering traveller for knowledge :
I shall then be satisfied, and not deem my time uiisspent if my labors tend to the public
?. NICHOLSON.
PREFACE TO THE REVISED EDITION.
Few words axe necessary to explain the present revision. When the work originally
appeared, it excited gixjat interest among artizans, and at once took the foremost rank among
works of the kind. By its intrinsic merit, it has ever since mainttiined this position,
although, latterly, many professedly new and original works have been issued, with the
avowed object of taking its place. They have failed to obtain foothold ; and the limited
sale which some few have been so fortunate as to secure, must he attributed to the extensive
and valuable contributions levied on this work of Mr. Nicholson's. Indeed, it has been for
many years the great source of supply to pseudo authors, who, having mutilated what they
took to avoid recognition, represented their alterations as new and superior designs. iSome
have, to a limited extent, succeeded ; but not among those who were acquainted with this,
the original and standard work.
Recently, however, such great advances have been made in the arts of Carpentiy and
Joinery, that many things in the work are now almost obsolete. Notwithstanding this, the
sales of the work were undiminished, showing the esteem in which it was held ; but the
publishers, unwilling that so valuable a treatise should be in any respect defective, determined
upon a revision. The present editors have undertaken it — with what success, the public
must judge.
All the plates have been re-engraved, and the matter they contained has been con-
densed, so that many valuable additions might be introduced, without increasing the bulk
of the work. A few plates, those on Practical Carpentry in particular, have been thrown
out, because they indicated methods entirely at variance with those in present use. A much
huger number has been substituted, comprising every principle of construction which the
carpentei' will need.
(V)
VI PREFACE TO THE REVISED EDITION.
Some years ago, a few plates on Stair-Casing were withdrawn, and improved methods
substituted by William Johnston, Architect, whose name appears on the title-page of the last
edition. All these methods of his have been retained entire, together with his remarks
upon them. Some plates, which were then rejected, have been replaced as too valuable
to be lost. To these have been added some original plates of Stair-Lines, in which the
|]ginciples are greatly simplified and may be applied to every kind of hand-railing. Many
other additions may be observed by inspecting the work.
The text also has been entirely re-written. A great part of it was so very obscure, as to
be almost unintelligible. This has been altered so as to be perspicuous ; and in some parts
the original explanations have been rejected entirely, and otliers introduced. It is not
claimed that the work is free from errors in this respect, but that it is free from all practical
error; and that the additions, substitutions, and alterations, will greatly improve its value,
and render it a complete compendium of Carpentry and Joinery.
THE EDITORS.
FniLADELTHIA, ./ff/^, 1853.
CONTENTS.
^^/WS<W^MM»/M^N/S^X^^^^^^^»<^^^^^^<^»^^*MI
PRACTICAL GEOMETRY.
PAtfB
Definitions '. 10
Problems 12
Drawing Instruments - 20
Mensuration 22
CARPENTRY.
Linings for So£Sts 25
Tracery 28
Arches 29
Groins 30
Niches 39
Roofs 43
Skylights 45
Domes 47
Practical Carpentry. — Designs 49
JOINERY.
Stair-lines (Johnston) • 68
Stair-lines (new) 76
Stair-lines (Nicholson) : 78
Diminishing Columns. 90
Sash-work .' ,,... 91
Architrave 7 : 92
Baking Monliiiai nMninus^ &c. 93
nber 97
(vii)
%
/
/-
V
PRACTICAL GEOMETRY.
VM>^M^MMMMm^>M»<M^^MM»^W^»WW»A^^«»^^V>/»^^^
Extension has three dimensions, length, breadth, and thickness.
Geometry is the science which has for its object, first, the measurement of extension ;
and secondly, to discover, by means of such measurements, the properties and relations of
geometrical figures.
Practical Geoicetry is that branch of the. science which describes, without demonstrar
tion, the various methods of constructing angles, figures, curves, and geometrical solids, and
comprises also the rules for mensuration.
C«)
PRACTICAL QEOMETBT.
PLATE 1.
DEFINITIONS.
A POUiT is that which bu Deither length, breadth, nor thickness, bat position 011I7.
A line is that which has length, without breadth or thickness.
A right or straight line preserves the same direction between any two of its points. See Fig. A.
A carve or carved line changes its direction at every point. See Fig. B.
A surface is that which has length and breadth, without height or thickness.
A plane is a surface, such, that if any two of its points be joined hy a straight line, that line will lie
wholly in the surface.
Every surface which is not plane, or composed of plane snrfaces, is a curved surface.
Two lines are said to be parallel, when, bang situated in the same plane, they will not meet, how far
uever, either way, both of them be produced. See Fig. C,
When two straight lines meet each other, their inclination is called an angle, which is greater or less
ftceording as the inclination is greater or less. The two straight lines are called the sides of the angle,
and their common point of intersection the vertex.
When one straight line meets another straight line, without being inclined to it on the one nde any
more than on the other, the angle formed is called a right angle, and the two lines are said to be perpen-
dicular to each other. See Fig. 2).
An angle less than a right angle is an acute angle. See Fig. S.
An angle greater than a right angle is an obtuse angle. See Fig. F.
An angle is designated by a single letter placed at the vertex, or by three letters, two of them upon
the sides, and the other at the vertex, the letter at the vertex being always placed in the middle, as the
angle c or aeb,in Fig. JS.
A polygon is a portion of a plane terminated on all sides by lines.
A polygon of three sides is a triangle ; one of four sides, a quadrilateral ; one of five, a pentagon ;
one of six, a hexagon ; one of seven, a heptagon ; one of eight, an octagon ; one of nine, a nona^tm
one of ten, a decagon.
An equilatefal triangle has all its sides equal. See Fig. G.
An isosceles triangle has two of its sides equal. See Fig. H.
A scalene triangle has all its sides unequal. See Fig. /.
A right-angled triangle has one of its angles a right angle. See Fig. J.
A trapezium is a quadrilateral which has no two of its sides parallel. See Fig. S.
A trapezoid is a quadrilateral which has two of its sides parallel. See Fig. L.
A parallelogram has its opposite sides parallel. See Fig. 3f.
A rhombus has its opposite sides equal and parallel — it« angles not right angles. See BSg. Si
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PRACTICAL GEOMETEY. 11
A rectangle has its opposite sides parallel, and its angles right angles. See Fig. 0.
A square has all its sides equal, and its angles right angles. See Fig. P.
A regular polygon is one whose sides and angles are equal each to each. Thus, Fig. 72 is a regular
pentagon, S a regular hexagon, T a regular octagon.
An irregular polygon is one whose sides and angles are not equal. See Fig. Q.
A polygon is said to be inscribed in a circle when the vertices of its angles lie in the circumference.
Thus, acdy Fig. X, is an inscribed triangle. The circle is also said to be circumscribed about the triangle.
A circle is a portion of a plane bounded on all sides by a curved line, every point of which is equally
distant from a point within, called the centre. See Fig. 27.
The radius of a circle is a right line drawn from the centre to the circumference, sa e d^ e a^ or
c 6, Fig. U.
The diameter of a circle is a line passing through the centre, and terminated on both sides by the
circumference, as a 6, Fig. ?7.
An arc is any part of the circumference, as a 6, Pig. Fl
A chord is a right line which joins the extremities of an arc, as a (, Fig. V.
A segment is the part of a circle included between an arc and its chord. See Pig. Fl
A sector is the part of a circle included between an arc and two radii drawn to its extremities. See
Fig. W.
A line is tangent to a circle when it has but one point in common with the circumference, and does
not intersect it. Thus, eah^ in Fig. X, is a tangent.
The circumference of a circle is divided into 860 equal parts, called degrees. Arcs are estimated
according to the number of these equal parts which they contain. Thus, in Fig. T^ if the arc d e contains
80 of these parts, it is an arc of 80 degrees, written 80^. If a & contains 90 parts, it is an arc of 90%
or a quadrant, that is, one-fourth of a circumference.
Arcs are also measures of angles, the vertices of the angles being supposed to be at the centre of the
circle. Thus, if the arc de, in Fig. Tj contains 80^, the angle dee is tin angle of 80^, and d e b is an
angle of 90^, or a right angle.
It will be observed that an angle of 45° is the half of a right angle. Thus, in Fig. Z, the angle fe e,
or fc ay is an angle of 45°. Also, an angle of 60° is two-thirds of a right angle. Thus, the angle gebis
an angle of 60°. The chord of 60° is equal to the radius of the circle. Hence, the triangle g e b iB
equilateral.
PRACTICAL GEOMETRY.
PLATE 2.
PROBLEMS.
Fig. 1. To draw a Perpendicular at a given Point in a Line.
Having taken two points, a and &, equally distant From c, the given point, aa centres, describe area inter-
Becting each other at d, and then draw dc; it will be the perpendicular required.
Fig. 2. From a given Point without a Line to let fall a Perpendicular upon the Line.
From the given point d, as a centre, describe an arc cutting the given line in two points, a and b ; with
these points as centres, describe two area intersecting each other at e ; then draw d e, and it will be the
perpendicular required.
Fig. 3. To biieet a given Line btf a Perpendicular.
From a and b, the extremities of the line, as centres, describe two arcs, intersecting each other at d
and e ; draw d e, and it will bisect the given line at e, and be perpendicular to it.
Second method. From a and b, aa centres, describe two arcs, intersecting each other at g, and from g
let fall a perpendicular upon a h.
Fig. 4. To draw a Perpf.ndic-ular to a Line at its Extremity.
Let b be the extremity of the given line ; from any point c above the line as a centre, describe the arc
abd\ draw acd and rf 6 ; it will be the perpendicular required.
Second method. Upon the given line take a h, equal to six units of measure, as inches, feet, or rods,
then, with a radina of eight units and centre b, draw a dotted arc above a 6, as at li, and with a radius of ten
units and the centre a, draw an intersecting arc at d; from d, the point of intersection, draw d b, and it
will bi; the perpendicular required.
Fig. 5. To draw a Perpendicular to a Line at or near its Extremity.
From the points a and g, as centres, describe two arcs intersecting each other at d and e ; draw d r, nrul
it will be perpendicular to the given line at c.
Fig. 6. Through a given Point to draw a Line parallel to a given Line.
Let c be the given point and a b the given line ; from a and o as centres describe the arcs a d au4 c h ;
make a d equal to b c, and draw nd; it will be parallel to ab.
Fig. 7. To bisect a given Angle.
From c, the vertex of the angle, as a centre, describe an arc a b ; from a and b, as centres, describe two
arcs intersecting each other at e ; draw c e, and it will bisect the angle.
Fig. 8. To make an Angle at a given Point on a Line equal to a given Angle.
Let y e be the line, y the given point, and/c j the given angle ; from c as a centre with any radius, as
e a, describe the arc a h ; from the given point y as a centre, and with the same radius, describe the arc x z
and make it equal to ab; then draw g z, and xgz will be the required angle.
Fig. 8. To divide a Line into Parts proportional to the Parts of a given Line.
Let y e be the given line, divided into parts at the points a, b,c,d; let y n be the line to be divided ;
join the extremities e n, and draw the tines d k, c i, &c., parallel to e n.
Second method. Let fg be the given line, divided at the points i and k, and d e the one to be divided.
From c, the vertex of the triangle cfg, draw the hnea c i and c k.
Fig. 9. Two Angles of a Triangle being given, to find the TJiird.
On the straight line a b lay oS bee equal to one of the given angles, and dee equal to the other ; and
« «■ «I will be the third angle required.
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PRACTICAL GEOMETRY.
PLATE 3.
PROBLEMS— (coMTUnriD).
Fig. 1. To eorulrwl
From c and h, the extremities of the given
J, ftnd draw ab, ae.
equilateral Triangle <m a given Line.
OB centres, and witb the radius e b, describe t
i interaoctitg
Fig. 2. T/tt three Siilet o/ a TriamjU heing given, to conslru^t the Triangle.
Make the eide ab equal to C, one of the given liuea. Frum the extremities a and h, with radii equal to X and B,
the other given liaea, describe two arcs intersecting at r, and draw a c and b c.
Fig. 3. To conttniet a Rixtanglc, equivalent Co a gionii Parallelogram.
Iiet abdc ht the given parallelogram. Coaatruct upon its base, c (f , a rectangle baviDg the same altitude as the
parallelogram.
Fig. 4. To coiatruet a Rhomhtu, tcilh a given Angle.
Hake the angle bad equal to the given angle. Make a h and a d equal, and draw b e punillel ia a d, and d c
parallel to ab.
Fig. 5. To conitruet a Sqtiare on a given Line.
From c and d, the eitremitiea of the line, as centres, describe the arcs ceb, and dea. From r set off « a, and e 6
equal to e/, the half of ec; then draw db, a h, und a c.
Sei'ond Method. At c und d, the eitremitieB of the given line, erect pcrpendi colors. From c and d, as centrea,
with the radius e d, describe area intersecting the perpcndiculurs at b and c, and ihen draw b c.
'^\g 0. To construct a Rectangle equivalent to a given Triangle.
At b and t, the CKlremities of the buse of the given triangle, erect the perpendiculars b d and e e. Intersect theae
perpendiculars b^ the line d r, purallifl to the base b c, und drawn through /, (be middle point of the altitude of the triangle.
Fig. 7. To conttruH a Square eqvira/ent to a given Rectangle.
Produce d c, one side of the rectangie, until c i, the part produoed, is equal to c b, the other side of the rectangla.
Bisect if t at n; from n, as a centre, with the radius n d, describe u semiciruumferonoe. Produce c & to r, and on c «
describe o square.
Fig. 8. To inicribc in a Circle a regvJar Hexagon and an cipiilnferal Triangle,
Apply the radius ca sis times to the circumference, and then will be inscribed a regubir hexagon. Joiu the oiler,
nate angtea of the hexagon, and there will be inscribed on equilateral triangle.
Fig. 9. To inicrihe m a Circle a regular Pentagon.
Draw two dinntctcts, o h and b i, perpendicular to each other. Bisect the radius be at c; take e d, equal to a «;
then from a, as u centre, aud with the radius a f^, describe the arc <//, und the chord a /will be one aide of the retiuired
pentagon.
ie in a Circle a Square and a regular Octagon.
at right angles to each other, and join their extremities. Bisect the aru
ind the choi^ a b, of half the are, will be the side of the octagon required.
Fig, 11. To make an Octagon out of a Square.
r the disgonals /e and d g ; from / and e, as ccutres, and with a radius equal to one-half of the diagonal,
res cutting the sides .of the square in u and b ; remove from each corner of the square a triangle equal taabg.
Fig. 12. Ta make a Stjanre equal to tieo given Squares.
Let b d and bi/he the two given squares. Having placed them so that the angle at b shall bc Tcrtical, as repro-
oeuted in the figure, join a and e, and conslruot a square on tt e.
. NiiTK. — The proposition upon which tlin aolution of the ahuve problem depends, vis., that in every right-angled triangle
tfae square upon the hvputhenusu is equal tu the sum of the squares upon tlio other two aides, iu of groat pructiea! ralue. A
new metliod of demonsiraiing Ibe propo'ilKin is sliown in Fig. 13, Lot a & c bo a right-anKld triangle. Pniduuc b a until
a 7 is equal tote; conslruot squares on b g, b a. nnJ ca; dtAw kh piirollul to/e. and c/kh will be a square on 6c; then
tlie whole figure, minus the four triangles, A, B, C, and D, is the square upon the hypulhenuao. The tinuie figure, minns
the tour equal triangies, C, D, E, and F, is the sum of the squares upon the uther two siilea; hence, the proposition ia true.
The solution of the problem in Plote 2. Fig. 4, depends, also, upon this proposition. The distAuces a b, and bd. are tha
base and perpendicular of n right-angled triangle, of which the distance a d is the liypotlienuBe, It will be observed Uia
•urn of the squares of 6 and 8, equut id 3Ij and 64, is equal to the square uf 10. which is 100.
Various other valuable praoticaJ results may be obtained from tlie dMnonstnUion, , A», for instance, the sqnnre on th«
hypothenuse is equal to twice the reotaugle ooniained by the sides, plus the square, upon the difference of the sides: all ef
wliich may be readily seen by a reference tu the figure.
The above demonstration of this oclebrated proposition has never before been published.— Ed.
Fig. 10. 5
To inscribe a square, dmw two c
subtended by one of the sides of the
describe a
t4 PRACTICAL GEOMETRY.
PLATE 4.
, PROBLEMS (Continubd).
Fig. 1. To find the Centre of a Circle.
Praw any chord, as ef. At i, the middle point of «/, erect the perpendicular a i, and produce it
^ J. Then c, the middle point of a 6, is the centre of the circle.
Fig. 2. To find the Centre of an Are.
I>raw two chorda, a i and a <2 ; bisect them by perpendiculars ; and <?, their point of intersection, is
^e centre of the arc
Fig. 8. To find the Length of an Arc.
Jjet ab e he any arc ; bisect it at 6, and draw the chord a b ; produce the chord a c, until it is equal
to twice a b ; again, produce e n, until n i is equal to one-fourth of c n; and a i is the length of the arc.
Fig. 4. To find the Length or Stretch-out of a Semieircum/erence.
Construct an equilateral triangle, m n t, on the diameter m n ; draw the tangent-line bad parallel to
m 11, until It meets t m and t n produced. Then i a (2 is the stretch-out of the semicircumference man.
Fig. 6. To draw a Tangent to a Circle at a given Point.
^ Let a be the given point ; draw bad perpendicular to the radius {; a, at its extremity, and it will be
the tangent required.
Fig. 6- -^ Tangent-line being given^ to find the Point where it touches the Circumference.
Take dy any point of the tangent-line, and draw dc to the centre of the circle ; on e d, bs 9k diameter,
describe a semicircumference, and the point a, where it intersects the circumference of the circle, is the
point of tangency.
Fig. T. To deserve a Segment of a Circle having a given Base and Height.
Let b d he the base, and af the height ; bisect b d hy the perpendicular a/ c, and draw ( a ; bisect ( a
by the perpendicular e e, and e, the intersection of the two perpendiculars, is the centre of the circle, from
which with the radius c a, the segment bad may be described.
Fig. 8. To draw a Segment by Rods to any Length and Height.
«
Make two rods, a b and a d, each being equal to the base b d ot the segment, to form the angle bad;
then, having them secured, and placed as in the figure, put a nail at b and one at d. Now, place a pencil-
point at a, and move the firame either way, sliding against the nails at b and <f, and the point a will mark
the arc of the required segment.
Second Method. — If the segment required is too large<<f be conveniently drawn in this way, we may
cut a triangular piece of board, as shown at I^. 9, the height t c of the triangle being half the height of the
segment. Now, by putting a nail also at a, we may, with this triangle, draw half the arc of the required
segment at a time, in a manner similar to the above, placing it, as shown by the rods, at e a and e b. The
augl^ a eb of the rods is equal to ft t d.
Fig. 10. To draw the Segment of a Circle by the Method qf intersecting Lines.
Let ft d be the base of the segment, and a 6 its height ; draw the chord a ft, and erect ft m perpendicular
K It, and e ft perpendicular to ft (2 ; divide ft 6 and 6 dj each into six equal parts, at the points 1, 2, &c. ;
divide, also, a m into six equal parts, at the pmnts 1^ 2^ &c, and draw the lines 1 1^ 2 2', &c ; and their
points of mtersection with the lines a 1, a 2, &e., are points of the curve; trace the curve through them, and
you will have the half-segment a ft. The oth«r half may be drawn in the aane way.
THE NEW YftP.K |
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PRACTICAL GEOMETHT.
PLATE 5.
OF SOLIDS.
DEFINITIONS.
A cube ifl a solid bounded by six equal fiquarea. See Fig. 1.
A prism is a Bolid whoae lateral faces are parallelograma, and whose upper and lower
bases are equal polygons. See Figa. 2 and 3, Plate 4.
A prism is triangular, quadrangular, or pentagonal, according as its base is a triangle,
quadrilateral, or pentagon.
The altitude of a prism is the perpendicular distance between its upper and lower bases.
Thus, a h, in Fig, 3, represents the altitude of the prism.
A prism is right when its altitude is equal to its side, or when the-eide is perpendicular
to the plane of tlie base; otherwise it is oblique.
A pyramid is a solid formed by several plane angles proceeding from a common point,
called its vertex, and terminating in a regular polygon, which forms its base. See Figs. 5 and 6.
A truncated pyramid, or the frustum of a pyramid, is that part of a pyramid which is
left after the upper part has been cut off by a plane parallel to its base. See Fig. 5.
The altitude of a pyramid is the perpendicular distance from its vertex to its base.
Thus, b c, in Fig, 6, is the altitude of the pyramid.
A cylinder is a solid generated by the revolnlion of a rectangle about one of its oides.
See Fig. 7. The side of the rectangle about which it is supposed to revolve, is the axis or
altitude of the cylinder. See a b, in Fig. 8.
A sphere is a solid generated by the revolution of a semicircle around its diameter.
See Fig. 9. The diameter of a sphere, is a line passing through the centre and terminating
on both sides in the surface. See a d, Fig. 10. The radius of a sphere is half of the dia-
meter, or a line drawn from the centre to any point of tlie surface. See b c, Fig. 10.
A cone is a solid generated by the revolution of a rightrangled triangle around one of its
sides. Thus, in Fig. 12, if the right-angled triangle be revolved around the side a b, it will
generate a cone, as represented in Fig, 11. In Fig, 12, a b is the altitude of the cone, and
b d the radius of the base.
If a cone be cut by a plane, making, with the plane of the base, an angle less than the
angle included between the side of the cone and its base, the section will be an ellipse. Thus,
the section formed by a plane passing through the line a b, in Fig. 11, is an ellipse. It is
also represented in Fig. 13.
If a cone be cut by a plane, making, with the plane of the base, an angle equal to the
angle included between the side and the base, the section will be a parabola. Thus, the section
formed by a plane passing through the line a c. Fig. 11, is a parabola. It is represented in
Fig, 14.
If a cone be cut by a plane, making, with the plane of the base, an angle greater than
the angle included between the side and the base, the section will be an hyperbola. Thus,
the section fonned by a plane passing through the line a d, in Fig. 11. is an hyperbola. It
18 represented in Fig. 15.
J
16 PRACTICAL GEOMETRY.
PLATE 6.
THE ELLIPSE,
DEFINITIONS.
An ellipse is a curve, Buch that if from any point two lines be drawn to two fixed points, their sum will
be always equal to a given line. Thus, let a. Fig. 1, be any point of the curve, and o and o^ the two fixed
points ; then, if o a + o' a is always equal to a given line, the curve is an ellipeie.
The two fixed points, o and o', are called focL
A diameter is any line passing throujgh the centre, and terminating in the curve.
The diameter which passes through the foci is called the transverse axis ; and the one perpendicular to
it is the conjugate axis. Thus, d e. Fig. 1, is the tranverse axis, and A b the conjugate axis.
PROBLEMS.
Fig. 1. To describe an Ellipse with a Stringy the Foei and transverse Axes being given.
Take a string equal to the transverse axis, and fasten its extremities at the foci ; then place a pencil
against the string, and move it around, keeping the string constantly stretched.
Fig. 1. To do the same with the Trammel^ the Centre and Axes being given.
Place the trammel at the centre, as seen in the figure, and so arrange the rod efg upon the arms, and
the pencil g upon the rod, that e g will be equal to the transverse, Kudfg equal to the conjugate axis. Move
the pencil around, and it will describe an ellipse.
Fig. 2. To describe an Ellipse by means of intersecting Lines, the Axes being given.
Describe a rectangle upon the axes, and divide the conjugate axis into a number of equal parts, at the
points 1, 2, 8, &c. ; divide the transverse axis into the same number of equal parts, at the points 1', 2% 8\
&o. ; then draw the lines A 1, a 2, &c., b 1^, b 2', &c., and their intersections will be points of the curve
Trace the curve thtough these points.
Fig. 8. To describe a rampant Ellipse.
This problem is performed in the same way as the preceding, except that the parallelogram a i s d is
used instead of the rectangle a 5 d c, in Fig. 2.
Fig. 4. The transverse and conjugate Axis of an Ellipse being given, to draw its representation.
Draw b s parallel and equal to A c ; bisect it at/, and draw a/ and b b, intersecting each other at h ;
bisect A ib by a perpendicular, meeting a b produced in c, and draw b c, meeting s c in e ; then from 6, as a
centre, describe the arc B h, and from c, as a centre, describe the arc A A;, and you will have one-fourth of
the curve. Draw the other parts in the same way.
Fig. 5. An Ellipse being given, to describe within it another, having the same Eccentricity, or the same
Proportions in respect to Length and Width.
Describe the rectangle abdc on the transverse and conjugate axis, and draw the diagonals a d and b c;
let a' b' be the conjugate axis of the required ellipse, and through a' b' draw a' V and </ d' parallel to D b ;
join a' cf and V d\ and jf ib! will be the tranverse axis of the required ellipse.
Fig. 6. An Ellipse being given, to find the Centre, Axes, and Foci.
Draw any two lines, a e and d e, parallel to each other, and draw t k through their middle points ;
biseoi ik at O, and O will be the centre of the ellipse.
From 0, as a centre, describe two arcs, intersecting the curve at m and n ; draw m n, and d c b, per-
poidicalar to ic, will be the transverse axis, and A c B, perpendicular to it, will be the conjugate axis.
From B, as a centre, witn a radius equal to O b, describe two arcs, intersecting the transverse axis, and
the points of intersection, and o', will be the foci.
Tig. & To draw a Tangent to an Ellipse^ at a given Point.
Lt't ^ be tibe gtv^i point ; draw { and i o^ to the foci, and produce o i to o ; bisect the angle o b o%
and tfatt bisecting line will be the required tangent
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PRACTICAL GEOMETRY.
PLATE 7. '
THE PAIIABOLA, HYPEHBOLA, AND CYCLOID
DEFINITIONS.
A pnrabok is a curve, any poipt of whith is equally distant from a fixed point and a given line. Let a b. Fig. 1,
b. tbe given line, and f the fixed point; then, for any point of the curve, as g, the dislauces O s und a c are equal.
TLc given line, A D, is called the directrix.
The fixed point, F, is called the focus.
The line, u n, drawn through the focuE and perpendicular lo A B, is culled the axis.
The line, m n, drawn through the focus, perpendicular to the axis, is called the parameter.
An hyperhola is a curve, in which the difTcrcncc of tno lioes, drawn from any of its points to two fixed points, is
constantly equal to a given line. Let B, Fig. 4, be any point of the curve, and a and r the two fixed points; then the
difference betmeeo A B and B P ia always equal lo a given lino.
The points A and F aro called the foci, and M the centre, of the hyperbola.
The two curves described around the foci A and F, are called branches of the hyperbola.
In common language, the term hyperbola is applied to a singlo branch of the curve.
A diameter is any line passing through the centre aud termiDatingon both Mdes of the curve. Thns, I L is a diameter.
The diameter K H, which, being produced, passes through the foci, is called the transverse axis.
The line N o, perpendicular to it, is the conjugate axis.
If a circle. E F D, Fig. 6, be rolled along a right line, A B, any point of tbe circumference, as D, will describe an
are, as d 6, which is called a cycloid.
The circle, E F u, is called the generating circle, and the point D, the generating point.
The right line, a b, equal to the circuiufereoco of the generating circle, ia called the base of the oydoid.
Tbe line, D E, is called the axis of the cycloid.
PEOBLEMS.
Fig. 1. To describe « Parabola.
Take a straight edge, A B, and T-square, a c ; fasten at o one end of a atnng, equal lo n o, and the other end at F ;
place a pencil against the string, keeping it always Etrclched, and move' the square along the straigbt edge. The pencil
will describe a parabola.
Fig. 2. To describe a Parabola hi/ iiiUrsecling Linet.
Take the rectangle XBca, and divide the sides a c and c H into lie same number of equal parla at the points
1. 2, 3, 1', 2', 3', &c. ; draw perpendiculars lo c H, at tbe points 1', 2', 3', &c,, and, also, the liuca A 1, a 2, &o., iatei^
aecling them ; trace the curre through the points of intersection.
Fig. 3. 3h do Ote mme by another Mdlml.
Tiike the triangle c e i^, and divide the sidee o c and d into the same number of equal parts at the points 1, 2, 3,
Ac. ; draw the lines 11, 2 2, 3 3, &a., and trace the curve so that these lines shall be tangent to it, as represented iu
the figure.
Fig. 4. To dacrile an Ifyperbola.
Fasten one end of a rod at a, and attach a string to tbe other end, at c; fasten the other end of the string at r ;
place a pencil against it, and move it round the point f, keeping tbe string always stretched.
Fig. 5. To da Ok tame by interuxti-ng Lines.
IKvide the sides a c and << B of the rectangle a b c a into tbe same number of equal parts at the points 1, 2, 8,
1', 2', 3', hx,. \ produce b A to 0, and trace the curve through the intersection of the lines 1, a 1, c 2, a 2, Ac
Fig. 6. To describe a Cycloid.
Upon OE, half the base of the cycloid, and EO, tbe radius of the generating circle, construct the rectongle o. eci-,
divide a e, and the semicircumference e r t>, into the same number of equal parts at the points 1, 2, 3, 1', 2', 3', &c., and
erect tbe perpendiculars 1' 1', 2' 2', Ac. ; from 6', as a centre, with a radius equal to c E, describe the arc C a ; from ft',
as a centre, with the same radiuD, describe the arc 5' i, and so on ) from the points 6', 5', &o., lay off, un these arcs, tlM
chords E 1, E 2, &c., and through their exUemitles, a, i, h, kc, trace tbe curve.
3
I
18 PRACTICAL GEOMETRY.
PLATE 8-
THE SECTIONS OP A SEMI-CYLINDER AND HEMISPHERE.
The section of a cylinder, made by a plane oblique to the axis, is an ellipse.
Every section of a globe is a circle.
Every section of a globe, made by a plane passing through its centre, is called a great circle, being the largest that
can be obtained by cutting the globe. Its radius is equal to the radius of the globe.
PROBLEMS.
Kg. 1. To find the Section of a Semi-cylinder, made hy a Plane at right angles to the Plane pacing through its Axis.
The section is a semi-ellipse, whose conjugate axis is the diameter of the cylinder, and whose transverse axis is the
intersection of the two perpendicular planes ; thus, taking ^ </, equal to e Oj the radius of the cylinder, as the semi-
oonjugate axis, and a b, the intersection of the planes, as the transverse axis, the semi-ellipse may be described with the
frammel, as represented in the figure.
Second Method. Divide the semioircumfcrenoe doc into any number of parts, and draw, from the points of divi-
mon, perpendiculars to c^ c, meeting a 6 at the points 1' 2' 3', &c. ; erect the perpendiculars 1' 1', 2' 2', &c., respectively
equal to the lines 1 1, 2 2, 8 3, &c., and trace the curve through V V, 2' 2', 3' 3', &c.
For, conceive the semicircle and the semi-ellipse to be turned around d c and a 6, so as to be perpendicular to the
plane abed, then will they occupy the same position as in the solid, and the lines 1' 1', 2' 2', &c., will be respectively
parallel to 1 1, 2 2, &c. ; hence, the semi-ellipse thus found is the true section of the cylinder.
Fig. 2. 2b find the Section of a Semi-cylinder, m^de by a Plane, forminy an acute Angle with the Plane passing
through its Axis.
In the right-angled ^triangle t' a/ if, at A, make the angle i' f of equal to the angle of the planes, and t' of equal to
the radius of the base of the cylinder; from s draw s x, parallel to cd; draw t x, equal to (f of &t A, and perpendicular
to ab; produce it until M* is equal to ^t' at A, and draw is; draw x g perpendicular to the tangent fg^ which is
parallel to c d; join e and g; draw the tangent Vh parallel to e^, and from Fdraw Vy perpendicular to cd, until it
Beets b a produced ; draw y m parallel to s i, and then draw s m perpendicular to s t and y m ; then the section will be
SB dUpae of which sm is the semi-transverse axis, and s4' equal to ef, the semi-conjugate axis. The ellipse may be
dcKribed with the trammel.
Saxmd Method. Draw 1 1, 2 2, 3 8, &c., parallel U) eg, and from the points 1, 2, 3, &c., in c d, erect the perpcn-
ifinkn 1 r, 2 2^, 3 3', &c. -, then, draw V V, 2' 2', 3' 3', &c., parallel to s i, making them respectively equal to 1 1, 2 2,
tt^ke.; tnee the curve through the points V, 2', 3', &c.
K^ t Tafimd the Section of a Segment of a Cylinder, made by a Plane forming an obtuse Angle with the Plane
of the Segment.
JfHrr the wgle i^ o^x, at B, equal to the angle of the planes, and draw aV perpendicular to a' x, and make it
ifmA r»jk.jim heght of the t^ment; draw bf <f perpendicular to aV; draw a b parallel to m n, and b c perpendicular
-t^^ / dHie^Mki Ui¥<f s^ B; produce c6 until c4' is equal to </ a' at B; join 4' and a. This line corresponds tu is,
Trc 1. imk cht cam naj he traced by the second method of the preceding problem.
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PRACTICAL GEOMETRY. 19
At Fig. 4 is exhibited the method of finding an oblique section of a solid of irregular form ; which is the same as
that employed in the second method of the problem preceding the last.
Kg. 6. Griveny the Position of three Points in the Circumference of a Cj/linder, and their respective Heights from the
Base, to find the Section of the Segment of the Ch/linder, through these three Points.
Let cedhe three points in the circumference of the base of the cylinder, immediately under the three given points,
and Z TXthe height of the given points respectively, above the base; join the points c and d, and draw ca, eA, and
d h, perpendicular to c d, equal to Z, Y, and JT, respectively; produce c d and a 5 to meet each other in 0; draw e D
parallel to d 0, and A D parallel to 6 ; join DO, In Z> 0, take any point, as D, and draw D H perpendicular to
D c, cutting Odin H] from the point H^ draw JJJ perpendicular io Oh, cutting it at K) from 0, with the radius Z>,
describe the arc /Z>, cutting JET/ in I, and join /; divide the circumference ced into any number of equal parts, and
from the points of division draw lines toed parallel to Z>, cutting cdin 1, 2, 8, &c. ; from the points 1, 2, 8, &c., in
c d, draw lines parallel to c2 5, cutting the line a 5 in l', 2', 8', &c. ; from the points 1', 2', 8', &o., in a 6, draw lines
parallel \jo Ij and make 1' 1' equal to 11 on the base of the cylinder ; make 2' 2^ equal to 2 2, 3' 8', equal to 8 8, &o.
Through the points V 2' 8', &o., trace the curvo; which will be the contour of the section required.
Fig. 6. To find the Section of a Hemi^here made hy a Plane perpendicular to its Base.
lieifd be the radius of the sphere, and a h the diameter of the section ; describe a semioircamference upon a 6, as
a diameter, and it will be the section required.
Fig. 7. To find the Section of a Semi^lipsoidy which is a Solid generated hy the Revolution of a SemireUips6
ahout its AxiSf made hy a Plane perpendicular to its Base.
Let cdhe the diameter of the circular base, and the curve codhe a section perpendicular to the base, and passing
through the centre ; then, let a h, parallel to c ^, be the diameter of the required section. The curve a (/ 5 may be
traced by means of the lines in the figure, according to principles heretofore developed.
At Fig. 8, is exhibited the method of finding any section of an irregular solid, generated by revolution. Let sfhe
the axis of revolution, c/the base of the generating figure, and a h the base of the section. The method will be readily
itnderstood by inspection of the figure.
PRACTICAL GEOMETRY.
PLATE 9.
DRAWING INSTRUMENTS.
The Geometrical Drawing of an object is usually made on a much smaller scale than tin
real size of the object. Thus, the drawing of a houee. for instance, is much smaller than tb<
hon3e. In order that the drawing may be a correct representation of the object, the relativi
size of the various parts of the drawing must be the same aa the corresponding parts of th^
object itself. In other words, the parts of the drawing must all be made on the same seals
There are a variety of instruments used in drawing, some of the most important of which s
represented in the plate.
Fig. I. This figure represents a drawing-board, a T-square, and a triangle. Drawing
boards are sometimes made with a frame surrounding the board, by which to confine thl?
paper; but experience proves that the kind here exhibited are cheapest and best. Anw
ordinary workman can make one. Take a board about two feet wide and three feet long, and!
screw cleats beneath, to keep it from warping. Dress the top perfectly smtroth and level, and
see that the ends are perfectly parallel and at right angles to the sides. To prepare it foj
use, cut your paper somewhat smaller than the hoard. Sponge it all over, and having paste
or glued the edges of tlie paper to the upper surface of the board, put it aside. In drying,!
the paper will stretch tight, and present a smooth firm surface to draw upon, which beinttl
done, the drawing may be cut out, and the board cleaned off with hot water.
The T-square is used to draw parallel lines. By moving the shoulder of the cross-piecaB
along the end of the board, any number of parallels may be produced.
In using the triangle, the T-square is held firmly on the board, and by placing tha'^
triangle, with either of its edges, against the rule of the square, oblique or perpendiculai
lines may be drawn to it, and by moving the triangle along the rule, lines parallel ■
these may be produced.
Fig. 2 is a flat instrument, usually made of Ivory, upon one side of which is a variefyl
of scales; those following the numbers 20, 25, &c,, are scales of equal parts, differing only inl
respect to the length of the unit of the scale. The first unit, which is usually not uumbered-i
is divided on the upper side into twelve, and on the lower side into ten, equal parts. Inl
drawing an object, having determined wliiclj of the scales shall be used, and how many units J
of measure of any given denomination on the object, as inches or feet, shall be represented |
by a unit of the scale, take from the scale, with the divisions, the number of units correspond-
ing to the length of any line on the object, and lay them down upon the drawing. The
fractional parts of the unit of the scale may be taken nearly by means of the divisions of thoi
first unit in the scale. When greater accuracy is required in the fractional parts, the diagonfd
scale of equal parts, which is represented on the lower side of the figure, is used. The fire
unit in this scale ia divided, both on the upper and lower side, into ten equal parts, am
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PRACTICAL GEOMETRY. 21
oblique lines drawn from the first point of division on the one side to the second on (he other,
from the second on the one to the third on the other, &c. It is, also, divided into ten equal
parts by horizontal lines. By this means, the length of a line can be taken off accurately to one
one-hundreth of the unit of the scale. If the fractional part of the unit consists of any number
of tenths and hundredths, open the dividers till one foot reaches the oblique line, denoted }fy
the tenths, counting from the right, on the horizontal line, denoted by the hundredths, count-
ing from the bottom.
The scale marked C HO is a scale of chords. It is formed, as the figure indicates, by
laying down, upon a right line, the chords of one, two, three degrees, &c., to ninety degrees.
This scale is useful in laying down an angle containing any given number of degrees, and in
mensiinng the number of degrees in a given angle. To lay down an angle containing any
number of degrees, draw one side of the angle ; from one extremity of the side, as a centre,
with a radius equal to the chord of sixty degrees, describe an arc ; lay ofF, from the side on
this arc, the chord of the given number of degrees, and, through the extremity of the chord,
draw from the centre the other side of the angle. To measure the number of degrees in a
given angle, from the vertex of the angle aa a centre, with a radius equal to the chord of
sixty degrees, describe an arc intersecting the sides, and the chord of the intercepted arc will
indicate the number of degrees in the angle.
The other side of thi.s instrument is represented in Fig. 4. By means of the graduation
of its perimeter, it is made to answer the same purpose as the semicircular Protractor, which
will soon be explained.
Fig. 3 is a Sectoral scale of equal parts. It consists of two arms, which turn upon a
joint, and upon each of which is a scale of equal parts. Its use is the same aa that of the
above scales of equal parts. Having determined upon a scale for drawing, as, for instance,
five feet to the inch, open the dividers to the distance of an inch, and then open the arms of
the instrument, until the dividers will extend from 5 on one arm to 5 on the other. Then,
to take off any number of feet with the dividers, as, for instance, nine, the angles of the arras
remaining unchanged, open the dividers until ita feet extend from 9 on the one side to 9 on
the other. Other scales are usually laid down on this instrument, but they are of no import-
ance to the student of practical carpentry.
Fig. 4 is a Semicircular Protractor. It is used, like the scale of chords, in laying down
and measuring angles. To lay down an angle of any number of degreeti, place the centre of
a circle at the vertex of the angle ; make the diameter coincide with one side of the angle,
and then count off on the circumference the required number of degrees, and draw from the
vextex the other side of the angle. To measure an angle, place the centre at the vertex of
the angle, and count on the circumference the number of degrees included between the sides
of the anglA>
MENSURATION.*
^^^^^^^^^^^^^^^^^^^^0^**^^^*^^*f^*^*^0*^**f*^^**0k^f^m^0t
OF SURFACES.
Wi deterTDine the area or contents of a surfacei by finding how many times the given surface contains some other 8iup>
hoe which is assumed as the unit of measure. Thus, when we say that a square yard contains 9 square feet, we should
onderstand th^t one square foot is taken as the unit of measure, and that this unit is contained 9 times in the square yard.
The most convenient unit of measure for a- surface is a square whose side is a linear unit in which the linear dimen-
sions of the figure are estimated. Thus, if the linear dimensions are feet, it will be most convenient to express the area
in square feet; if the linear dimensions are yards, it will be most convenient to express the area in square yards, &o.
I. To find the Area of a Square, a Rectangle, or a Parallelogram.
MuUtply the base hy the altitude, and the product will be the area.
Ex. 1. What is the area of a square whose side is 204*3 feet? Ans, 41738*49 sq. ft.
Ex. 2. To find the area of a rectangular board, whose length is 12} feet, and breadth 9 inches. Ans. 91 sq. ft
Ex. 8. To find the number of square yards of plastoring in a parallelogram, whose base is 87 feet^ and altitude
5 feet 8 inches. Ant, 21 j^ sq. yds.
[. To find the Area of a Triangle,
Multtpljf the hate by the aUihtde, and take half the product Or, multiply one of iheu dimensione by haJlf the gQwt.
Ex. 1. To find the number of square yards in a triangle, whose base is 40 feet^ and whose altitude is 80 feet.
AnM, 66f 8q.> v ^
nL To find the Area of a Trapezoid.
Add IkN/eAa* the two parallel tides; then multiply their turn by the altitude of the trapezoidy and half theproduU
wHl be the required area,
Ex. 1. How many square feet are contained in a plank, whose length ift 12 feet 6 inches, the breadth at the greater
end 16 inchesi and at the less end 11 inches? Ant, 13^] sq. ft.
tV. To find the Area of a Quadrilateral
Join two of the angkt by a diaganaly'' dividing the quadrilateral into two trtanglet; then, from each of the other
anglety let fall a perpendicular on the diagonal; then multiply the diagonal by half the turn of the two perpendiculart^
and the product win be the area, *
Ex. 1. How many square yards of flooring in the quadrilateral, whose diagonal is 65 feet, and the two perpendiculars
let faU upon it, 28 and 88} feet? ^im. 222y^ sq. yds.
V. To find the Area of a regular Bslygon,
Multiply half the perimeter of the polygon by the apothegm or perpendicular, let fall from the centre am one of itt
tidet^ and the product will be the area requirfd,
Ex. 1. To find the area of a regular hexagon whose wdes are 20 feet each, and whose apothegm is 17*3205 feet
.4f«. 1089.28 sq. ft
* Davies' Legendre.
(22)
MENSURATION. 28
VL To find the Circumference of a Circle when the Diameter isgiveny or the Diameter toAen the Circumference is given.
Multiply the diameter hi/ 8*1416, and the product will be the circumference; or, divide the circumference hy S*1416|
and the quotient toill he the diameter.
Ex. 1. What is the circumferenoe of a circle whose diameter is 25? * Am. 78*54.
Ex. 2. What is the diameter of a circle Tshose circumference is 6850 ? An$. 2180*41.
VII. To find the Length of an Arc of a Circle containing any Number of Degrees.
Multiply the number of degrees in the given arc hy 0-t)087266, and this product hy the diameter of the circle.
Remark. — When the arc contains degrees and minutesi reduce the minutes to the decimal of a degree, which is
done by dividing them by 60.
Ex. 1. To find the length of an arc of thirty degrees (SO''), the diameter being 18 feet Ans. 4-712864 ft.
VIII. To find the Area of a Circle.
Multiply the circumference hy half the radius. Or, multiply the square of the radius hy 8-1416.
Ex. 1. How many square yards in a circle whose diameter is 8| feet f Ans. 1*069016.
IX. To find the Area of a Sector of a Circle.
Multiply the arc of the sector hy half the radius.
Ex. 1. To find the area of a sector whose arc is 20 feet, the radius being 10 feet Ans. 100 sq. ft.
*
X. To find the Area of a Segment of a Circle.
Find the area of a sector having the same arc ; find the area of the triangle formed hy the chord of the segment
and two radii of the circle ; then add these two areas together when the segment is greater than a semicircle, and subtract
the triangle from the sector when it is less.
Ex. 1. Required the area of the segment whose cord is 16, the diameter being 20. Ans. 44*764.
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OF SOLIDS.
The Mensuration of Solids is divided into two parts : first, the mensoiation of their snrfiuies; and, secondly, the
mensuration of their solidities.
ft
We have already seen, that the unit of measure for plane surfsices is a square whose side is the unit of length. A
curved line, which is expressed by numbers, is also referred to a unit of length, and its numerical value is the number
of times which the line contains its unit. -If, then, we suppose the linear unit to be reduced to a right line, and a square
constructed on this line, this square will be the unit of measure for curved surfaces.
The unit of solidity is a cube, the face of which is equal to the superfioial unit in which the surface of the solid is
estimated, and the edge is equal to the linear unit in which the linear dimensions of the solid are expressed.
The following is a table of solid measures : —
1728 cubic inches = 1 cubic foot
27 cubic feet = 1 cubic yard.
4492i cubic feet = 1 cubic rod.
I. To find the Surface of a right Prirnn.
Multiply the perimeter of the base hy the altitude, and the product wiU be the convex surface; to this add the
area of the ttoo bases, when the whole surface is required.
Ex. 1. What must be paid for lining a rectangular cistern with lead, at 2d. a pound, the thickness of the lead
being such as to weigh 7 ^. to each square foot of surface; the inner dimensions of the cistern being as follows^ via
the length 8 feet 2 inches, the breadth 2 feet 8 inches, and the depth 2 feet 6 inches f Ans. 21. Zs. 10|cf.
84 MENSURATION.
n. Tojind the Surfiux of a right Pj/ramid.
Maltipls the perinteter of the bate if half the dattt height, and ikeprodwt will be the convex nir/aee; to thu add
the area of the bate.
Ex. I. Wh&t is tbe entire snr&oe of k rigbt pynmid whose slant height is 16 feet, and the base a peotagon whose '
udes are each 25 feet f An*. 2012-798 sq. ft.
in. ShJindOu Solidify of a Fi-itm.
Find the area of the bate; multiply ihit area bg the altitude, and the product vriU be the toUdify of the prism.
Ex. 1. What are the solid oooteDla of a cnbe whose side is 24 inches 7 Am. 1S824 en. tn.
Ex. 2. How man; cubic feet in a block of marble of which the length is 3 feet 2 inahes, breadtb 2 feet 8 inches,
aad height or thickness 2 feet Q inchesT Am. 21J cu. ft.
Ex. 8. Required tbe solidity of a triangular prism whose height is 10 fee^ and the aides of its triangular base S, 4,
and 5 feeL Am. 60 en. ft.
rV. Tojind the Solidily of a I^p^mid
Mvitiply the area of the bate hy one^hird of the altitude, and the product wHl be the toliditi/.
Ex^ 1. What is tbe solidity of a pyramid, each side of its square base being 80 feet, and the altitude 26 feet t
Ant. 7500 on. ft.
T. Ihjind the Surface of a Cylinder.
MuUi^y the cireumferenee of the bate by the altitude, and the product leill be the convex turfaee; to ihit add the
areat of the two batet.
Ex. 1. Required the entire snr&oe of a cylinder, the diameter of whose base is 2, and whose altitude ia 20.
Ant. 131-9472.
VI. Tofndthe Surfaceofa Gaie.
Jdultiply the ciTcumference of the bate by half the tfant height; to thit add the area of the bate.
Ex. 1. Required the entire surfaoe of a cone whose slant height ia SS feet, and the diameter of its base 18 feet
Ant. 1272-348 sq. ft.
VII. Tofnd the SoUdUjf of a Cylinder.
Multiply the area of the bate 2y the attitude.
Ex. I. Required the solidity of a cylinder whose altitude is 12 feet, and the diameter of its baae 15 feet
^n«. 2120-58 ou. ft.
VIII. To find the Solidity of a Cone.
MuJiiply the area of (he bate by the altitude, and take one4hird of the product.
Ex. 1. Required the solidity of a cone whose altitude ia 27 feet, and the diameter of tbe base 10 feet
Ant. 706-86 en. ft ,
IX. To find the convex Surfaoe of a tpherical Zone.
Multiply the altiiude of the gone by the circumfireaee of a great circle of the tphere, and the product utiU be th
convex turfaix.
Ex. 1. The diameter of a sphere being 42 inches, what is the convex surface of a xone whose altitude is 9 inches 1
Ant. 1187-5248 sq. in.
X. Tofvid the Solidily of a Sphere.
Multiply the turfaee, found by &e preceding rule, by one-third the radiui. Or, cube the diameter, and multiply
the number thut found by 0-5286.
Ex. 1. What is the solidity of a sphere whose diameter is 12 f Ant. 904-7808.
XL To find the Solidity of a tpherictd Segment.
Find the areat of the two baiet, and multiply their ram !>y half the height of the tegment; to thit product add At
totidity of a ^ihere whote diameter it equal to the height of the tegment.
Remakk. — Wben tbe segment has but one base, the other is to be considered eqnal to 0.
Ex. 1 What ia the solidity of a spherical segment, the diameter of the sphere being 40, and the distances from the
centre Ic the bases being 16 and 10 T Ant. 4297-7088.
mmr*^mm
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•; i I ..►.•: f o U ts DA 1 ION
fry"'"''"'' -Av-
or CARPENTRY.
LININGS FOR SOFFITS.
DEFINITIONS.
The Lining of a Soffit, in Carpentry, Bigoifies the covering of any concave Burface. The
lining is drawn aa if spread out on a plane, so that when bent to the curve it will esactly
fit the Soffit.
A Soffit, in Arcliitecture, is the under side of the head of a door, window, or the intrados of
an arch. It may be either plane or curved.
PLATE 10.
Fig. 1. Tu dram the Lining for Oie Soffit of a Windmc or Door, Jutving parallel Jambs and a
semicircular Head, which cuts oUi^juely Uiroiujh a straight Walt,
Let Che the plan or opening of the window, and let the base of the semicircle B be
drawn at right angles to the jarabs or sides of the plan C. Divide the semicircu inference
into any number of equal parts, as ten, and from these points draw perpendiculara to its base
across the plan ; extend the parts around B on the stretch-out line, and from these points
erect perpendiculars ; now make 1' a' ^ 1 a, 2' i' = 2 b, and so on. Trace a curve through
the points a' 1/ & d', &c., and it will be one edge of the lining; the other is obtained in a
similar manner.
If it is desired to make the cylinder, whose length shall be only the thickness of the
wail, the form of the end of it is seen at i>, which is of course a semi-ellipse, traced by
ordinates from the semicircle B.
Fig. 2 shows the method of getting the lining for the soffit of a semicircular window or
door-head, cutting right into a circular wall. The principle of this, and the following, is
similar to the preceding, and requires no further explanation.
Fig. 3 shows the lining for a soffit of an opening, which cuts obliquely mto a circular vail
4 v:263
IININGS FOR SOFFITS.
PLATE n.
Fig. 1. To draw the Lining for the Soffit of an Opening, with a circular Head splaying eqm
all around, in a straight Wall.
Let A be the plan of the opening; continue the sides or jambs a c and d b until they '
meet in e; then, with the centre e, describe the edges of the lining C; make the curve a If
equal in length to the semicircumlereuce at B, and the lining C, of the soffit, will be
determined.
For, conceive the Bemicircle 5 to be turned at right angles to the plan A, then every
point in the eemicircumference at B will be equidistant from the point e. But Cis described
with the same radius = a e\ therefore, the edge a h' of the lining (7, may be brought to
coincide with the arch of the semicircle B.
Fig. 2, To draw the Lining for the Soffit of an Opening, with a circular Head splaying equally
all around, in a circular WaU.
Divide the semicircumference at B into any number of equal parts, as ten, and draw
perpendiculars from these points to the base oo; thence draw lines to meet in /; from the
points 1' 2' 3' 4', &c., in the stretch-out line, which correspond to 1 2 3 4, &c., at B, draw lines
also converging to f; now, from the points of intersection a b c d e, on the plan of the opening
A, draw lines to of, parallel to the base o o, and with the centre f, and radius/i, draw the
arc ie'; also, with/A, draw the arc kd', and bo on; then, by tracing a curve through the
points a' b' (/, &c., we have half the outer edge of the lining, from which the other half is
easily obtained. The lower edge is found on the same principle, and thus the lining Cof the
soffit is determined.
For it is easy to conceive that, if the semicircle B be turned at right angles to the plan
A, the points 1' 2' 3', &c., may be brought to coincide with the [mints 12 3, &c., on the arch
B; then, the points a' b' c*, &c., will stand perpendicularly over the points abc, &c., on the
plan ; because the arcs ea',fh', gc', &c., wilt lie directly over the lines e a, f b, g c, &c.
Note. Tbe leanier is adfised to cut this, and tbo following soffits, out of pasteboard, which will fRtniliarise tha
nrinciplea
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OMJ
LININGS FOK SOFFITS.
PLATE 12.
Fig, 1. To draw the Linmg of a cyUndrical Soffit cutting right in a Wall which does not stand perpendicular
to the Ground, so that the Edge of the Lining will Hand directli/ over the base Line of the Aperture.
Let a g, at A, be the level of the ground, and a m the line of the wall, eqaal to the radius of the cylin-
der ; with this railiua describe the Bemicircle at B ; take the distance ab to the foot of the perpendicular m b,
at A, and place it from a to 4, at B, Now, upon the lines o a and a 4, ae axes, draw, by means of the inter-
eecting lines exhibited, the semi-ellipse 0, 1, 2, 3, 4, &o. ; divide the semi circumference into any number of
equal parts, as four, and let fall the perpendiculars to the base line o e ', also, at C, make the line o o equal
to the stretch-out of the semi circumference, at B, and lay down the corresponding division. Now, at A, draw
mg perpendicular to m a, and take from B the distances if 1, r? 2, i 3, a 4, and set them, at A, from j, thus,
gffe, ed, dc\ erect perpendiculars from each of the points/, e, d, c, towards the hne^m; on the lining,
0, make the distances Id, 2 c, 3 J, 4 o, equal to the distances g h, h i, i k, k I, at A, respectively ; then,
trace a curve, at C, through the points o, d, e, h, a, and these points of the lining, when it ia hent around the
curve on the plan B, will atund directly over o, d, e, b, a, at B, and the points 1, 2, 3, 4, of the lining C, will
coincide with the points 1, 2, 3, 4, at B.
Fig. 2. To find the Lining for the SofU of an Aperture in a straight Wall, whose Plan it a Trapezoid,
and whose Elevation on the inside is a Semi-ellipse, and on the outside a Sctnictrcle, to related to each
ether, that a Straight-edge coinciding with the lined Surface may everywhere be horizontal.
Let the trapezoid a b c D be the plan of the aperture, A B being the inner side, and c D being the outer
side, while a C and B D represent the jambs ; with the centre Q and radius g d describe the semicirde D e' C. which
will be the outside elevation ; also, taking E F ^; s', describe the semi-ellipse ABB, which will be the inside
elevation ; continue the jamhs A o and B D until they meet in H, and through B and o draw the line u H.
Now, divide the quarter circumference C b' into any number of equal parts, as five, and from these points
1, 2, 3, 4, let fall perpendiculars to a. From H, through the points e,f, g, h, the foot of each perpendicular,
draw lines intersecting the line A B.
Draw the line H K perpendicular to A H, and equal to c Q, and divide it like to c a, making H i =: a A,
Im, hg, &c. Now, with the radius h e and the centre i, describe an arc towards «' ; also, with the radiuu
ii/and the centre m, describe an arc towards/', and so on. Having taken the fifth part of the arc c b',
such as 1, fix one foot of the dividers in c, and with the other intersect the first arc at e' ; then, place the
foot in e', and intersect the next arc at/', and so on, to g'. Trace a line through the points c, e', /', g',
h', q', and this will be half the outer edge of the lining, which will lie over the line c, «,/, g, k, a, of the plan,
and will coincide with the arc c, 1, 2, S, 4, e', which is turned around c s, so as to stand at right angles to
the plan. The other half of the outer edge of the lining o' d' is similar, and found by inversion, making
the angle i e l := i s h, and the divisions on s L equal to those on K H.
Extend the lines I e', my, n g', and o h', towards «', h', e', d', and take e* a' ^ e a on the plan ; th%
f h' ^fb, and 80 on. Then, through the points A, a', b', c', d', f', trace tho line A f', and, ha\'ing obtij i-d
in like manner its prolongation f' b' the line a e' will be the outer edge of the lining. Now, the figure A p'
D' will be the entire developmebt of the lining, wtucb, when bent over the plan, will exactly fit the soffit.
28 LININQS FOR SOFFITS.— TBACEBY
PLATE 13.
Fig« 1. lb find the lAning Jar the Soffii of an Aperture in a circular WaUj (he Jamba being
inclined to eocA other^ and the covered Surface splaying^ so thai a Straight^ge^ coinciding
with ii^ mag everywhere be horizontal.
The method for obtaining these lines is similar to that es:hibited on the last plate. The
nature of the aperture is the same, and the principle explained there, as applying to a straight
vrally is here applied to a circular wall. The curve of the wall is drawn upon the plate with
a very short radius, in order to exhibit the method more plainly. Find the line « g' c/, as in
the last plate ; take the distances firom the line sGoto the arcs representing th^ faces of the
wall, and trauafer these distances to the lining, and the edges will be found as in the last
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TRACERY.
Fig; 2« Smng given ang QoAic Archy to draw another^ eiAer right or rampanty so thai the ttoo
^mU inkretci or mitre trulg together.
Let A be the given arch* Draw the chord a o« and divide it into any number of equal
parta^ a;» ibar ; tken« iVom the pcdnt f, draw lines through 1« 2^ 3, intersecting the arch at h gf.
JBrect a d perpendicular to a <^ and from o, through fg hy draw lines intersecting the perpen-
dicular in %1y Cy 6. Now, let the arch B, which we wish to draw in correspondence with A^
have the same heights e ^ and a greater ba$e> a e ; draw the line a o, and divide it also into
four equal parts ; mak» the divisions on a <f equal to those on a d at J^ and draw lines firom o
to the points byCyd. Having drawn Une$ from e, through 1« 2^ 3, trace the curve o/g h Oy
through the points of intersectioa. This will give the deeined airh.
£v similar constractioDiy the rampant arch C may be made to corre^Kmd
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PLATE 14.
7b draw Arches of varifniB Forms, and to find t/ie LtJiee of the Jointa betuKen the Archrstonea,
Aa carpenters are frequently called upon to prepare centring for arches, and also to cut
out patterns of the arch-stonea, to be used by the stone-cutter, we have thought best to intro-
duce a plate wliich will familiarise the student with the beat methods of drawing and dividing
arches.
Fig. 1. This is the semicircular, or perfect arch. It is drawn from the centre c, and
the joints between the voussoirs are parta of the radii. If it is not convenient to draw the
radii, make each line perpendicular to a tangent, as at t.
Fig. 2. This is a diminished, surbased, eegmental, or imperfect arch ; being composed
of an arc less than a semicircle. An easier method of drawing the joint is here exhibited.
From the points 1 and 3, as centres, draw small intersecting arcs above the arch ; and from
the point of intersection, draw a line through the point 2, bisecting 1, 3 ; this will give the
line of the joint correctly.
Fig. 3, This is an Arabesque, Saracenic, or horee-shoe arch, sometimes also called the
Oriental arch. It consists of an arc greater than a semicircle. The joints between the
springers lower than the centre, must not be drawn from the centre, as c c, but must be
made parallel to the imposts, or base line d e, as a b.
Fig. 4. The elliptical arch. Various methods for drawing the ellipse are laid down in
the first part of the work, and it is unnecessary to repeat them. To draw the joints, let F f'
be the foci of the ellipse; then, a line bisecting the angle f 1 f' will give the first joint, and
so on. If the curve is to be composed of a series of arcs of a circle, the points o, a, h, c, may
be used as centres.
Fig. 5. The Gothic lancet arch consists of two arcs, the radii of which are longer than
the span A b. The joints are drawn from the points o &.
Fig. 6. The equilateral arch is described by radii equal to the span.
Fig. 7. The obtuse pointed arch is described by radii shorter than the span.
Fig. 8. The ogee, contrasted, or reflected arch, is described from four centres, two
within, and two without the arch, a, b, o, o'. The proportions may be varied at pleasure.
Fig. 9. The Tudor arch is described from four centres within the arch, o, o', o, o'. For
an arch whose height is half its span, they may be found thus : divide the base line A u into
three equal parts at o and o'; then will o' be the centre of the arc an; from d, through
0, draw a line, and make the distance from c to o' equal to c d; then is o' the centre with
which to describe the arc d n. Those arches which have their height greater or less than
half the span, are found by other rules; but the proportions here given are perhaps the best.
The joints between the voussoirs, in this case also, lie in the direction of the radii.
\
80 OBOINS.
PLATE 15.
GROINS.
DBFINITION.
Gbouss aore formed b^ the intersection of arches or Taolts, and the Borfaoes where they meet may be
oonmdered as the sections of cylinders, &o.
DESCRIPTION OF THE CENTRING FOR BRICK-GROINS.
Fig. 1. P, P, P, P, 18 the plan of the piers which the vault is to stand upon ; a 6, Fig. 2, is the end
opening, which is a given semicircle ; and beiB the opening of the side-^urch, which is to come to the same height
as the end-arch a i : fix your centres over the body-vault. Fig. 1, as shown in the section at (7, then board
them over. In Fig. 1 is the manner of fixing the jack-ribs upon the boards, which is likewise shown at (7.
To find the Mtmldfor the Jack-ribe.
Take the* opening of your arches in Fig. 1, that is, a h and a d^ and lay them down in Fig. 2, at a i and
h <?, to make a right-angle. Divide one half of the given semicircle, J?, into five parts, and draw perpen-
diculars across, at 1, 1, 1, &c., to cut h d and dc^ the diagonals in 2, 2, 2, &c. Now, through the points
2, 2, 2, &c., draw lines parallel to 1, 1, 1, &c., at the base of J?, both ways towards F and (7 ; stick in nails at
1, 2, 3, 4, 5, in the arc of E^ and bend a thin slip of wood round them, which mark with a pencil at every
nail ; this slip of wood being stretched out from d, 1^ 2', 8', 4', 5', and perpendiculars drawn towards Q-j
will intersect the other lines, making small rectangles : a curve being traced through the intersections, will
give a mould to set the jack-ribs.
To place the Jack-ribe.
Bend your mould, (7, from a to the crown at €, in Fig. 1, which will give the edges of your boards ;
then fix a temporary piece of wood, level upon the crown, in the direction of //, and let it come the thick-
ness of the boards lower than the crown, and it will give the height of the jack-ribs, which is a very sure
method of placing them.
To find a Mould to out the Ends of the Boarde.
The semi-ellipse F is traced to the height of E, or drawn by a trammel. Take the parts round P, and
lay them out to 1', 2', 8', 4', 5' ; then H will be found in the same manner as Gy which will be a mould to
out the ends of the board that goes upon the jack-ribs against the body-vault.
To find the Moulde when both Arches have the same Opening.
Fig. 8. Take half the opening of the arches, whatever they are, and draw a quarter-circle, and d vide
H into riz parts ; bend a slip round it to take its parts, then stretch it out upon the base from to 6', and
draw perpendiculars from the points 1', 2', 8', &c. Through the points in the arc, draw the lines on bxah
^<s, p^raOel to 0. 6' ; the curve beinff traced as before, gives both moulds of an equal and similkrcfonn.
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PLATE 16.
The Plan and Inclination of an ascending Qrmn, one of ike Body-ribs, and the Place of the
Interaection on the Plart, being gwen, tojtnd the lihrm oft/ie Side^-ihe, ao that the Inkraectum
of the Arches shall lie in a perpendicular Plane,
Divide half the circumference of the body-rib, at B, into any number of equal parts;
draw lines from these points perpendicular to its base, and continue them to the line of
intersection on the plan; from thence, let them be drawn at right-anglea towards C, and
make the distances \h,\c,\d,\e, &c., at G, equal to the corresponding distances at B; then
■will the curve abed, &c., be the true curve of the side-rib. This curve is a semi-ellipse, and
may be found by intersecting lines, as at F, according to the rule for describing a rampant
ellipse.
7b find the Moulds for jtlitcing (he Jack-ribs.
At c, draw lines from the points a, b, c, d, &c., perpendicular to the line of ascent h g,
towards D and E; draw the semi-ellipse A as wide as the body-range, and as high as a A at
c; continue the ordinates \b,\c,\d, &c., up to 4; bend a slip around A, and mark upon
it the points 0, 1, 2, 3, &c. ; extend the slip upon the line k&, bX D and E, and divide it
correspondingly; now, through each of these points, draw perpendiculars across i 6 to inter-
sect the lines drawn perpendicular to the rake; then, curves traced through the points of
intersection will give the moulds for placing the jack-ribs. The edges of these moulds, bent
over the body-vault when boai-ded in, will exactly coincide with the intersection of the aide
and body vaults.
7b find the Jael-ribs of t/te Side-groins.
Draw the number of the jack-ribs upon the arch B, at their proper distances, and take
their fieveral heights, hi, kl, m m, &c,, and set them upon the arch G from a to 6, from b to
c, from c tod; draw lines through bed, parallel to the rake, and they will show on the curve
the proper length and form of the jack-ribs.
Tfa bevel the Body-riba.
Since all the body-ribs stand perpendicular to the plan, the upper edge must be bevelled
to correspond to the rake of the groin. To do this, let the under edge 1 I 1 1, at .^ of the
body-riba, be bevelled according to the rake, so that they may stand perpendicular ; then,
take a mould from B, or one of the body-ribs will answer, and place it on each side of the
rib to be cut, making the lower bevelled edges correspond. The upper edges may now be
raark*"* ""d bevelled.
PLATE 17.
T!ie ttDo Arches and Inclination of an ascending Qrmn being given, Co find (lie Place of the
IiUersection on the Pian.
Divide half of the body-rib' -B into equal parts, and thence draw lines, parallel to its base,
to the line a/. With m aa a centre, draw the concentric arcs from the points b, c, d, e,/, around
to h, g, iy k, I, and thence draw lines, parallel to tlie rake, to cut the arch C at 1, 2, 3, 4, and
a, h, c, d. From these points, draw lines through the plan, perpendicular to the direction of
the body-vault. From the centre g, erect a perpendicular to the rake, cutting the arch at 6,
From 5, draw another line through the plan. Also, from 1. 2, 3, 4, 5, at B, draw lines through
the plan, parallel to the direction of the body-vault, and trace curves from the piers to A,
through the points of intersection. These curves will be the place on the plan of the inter-
section of the vaults.
The moulds D and E, and the jack-ribs, are found in the same way as those on the last
plate.
Remarks. — Groins similar to these last may be seen under the Adelphi Buildings in the
Strand, London, where the declivity in going to the river is very steep.
In practice, there is no occasion for tracing these intersections on the plan, since the
moulds D and E are sufficient, without reference to the plan. To use these moulds, for this
and all otlier kinds of brick-groins, the centres or body-ribs must be placed first, as if there
were no side-arches cutting across them, and then closely boarded over. The moulds D and E
must both be bent over the body-vault together. Place the points I and e upon the piers at
and e. Then, keeping the top-points at 5' together, bend the moulds over the body-vault,
and the point 5' will stand perpendicularly over h on tlie plan. Now, along the inner edges
of the moulds, draw a curve on the boards of the body-vault, and this will be the line of
intersection. The jack-ribs of the side-vault are set to this hue at the proper distances from
each other.
It must further be observed, that the arch B must not be used instead of the arch A,
since this would produce a very great error in the moulds D and E; for it is evident that a
section of the body-vault, perpendicular to its axis, must be less in height than the vertical
section B, since this is oblique to the cylinder. Hence, the inoidda being drawn by means
of perpendiculars to the rake, the arch A must necessarily be used in finding them. If all
these things are properly understood, do difficulty can occur in brick-groins which may not
easily be surmounted.
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GROINS. 38
PLATE 18.
Fig. 1. The DiagonaJs of a plaster Oroin, which are straight upon the PUmy amd one \jf (he
Side-arches being given^ to fimd the Jachrihe.
Lay down the plan of the ribs, as at B^ and draw a rib upon each opening ; draw per-
pendicular lines from the plan of each opening, at the extremities a c «, to cut its corresponding
ribs eXhdfi then, the distance £rom i to i shows the length of the first jack-rib, from dXod
the length of the second, and from / to / the third.
How to bevel the Angle^ibs, so that (hey shall range wUh the Openiny of the Qroin.
First, get out the ribs in two halves, or thicknesses, as at J^ and F\ then, draw the plan
of the angle-rib, which, placed between E and Fj will show the true ranging upon the bottom
i>f the rib ; now shift your hib-mould parallel to the base of E and jF", and it will show how
much wood there is to be bevelled ofi*; then nail the two halves together, and the rib will be
complete.
lb find (he Jach-ribsj when (he given Arch is the Segment of a Circle.
The ribs in this case may be found by the method explained on Plate 13, Fig. 2, as
shown upon this plate at B, Ej and Fj Fig. 2 ; also, we may take the height of the segment
Ay Fig. 2, and place it fix)m btocj at G and D ; now take twice the radius a c, at A, and
place it from c and c, the crowns of O and 2>, to a and d ; the arches G and 2>, which are
parts of ellipses, may then be drawn by intersecting lines, as explained on Plate 6, Fig. 2.
Either of these methods is much easier, in practice, than to trace the ribs through ordinates.
34 GROINS.
PLATE 19.
Oiven one of Uie Body-^ibe, the Angles straight upon the Pla/n^ and the Ascent of a Oroin not
standing upon level Oroundf to find the Form of the ascending Arches j a/nd the Angh-ribs.
Let & a c at ^ be the angle of the ascent ; from the poin^t b make b c perpendicular to a b,
and describe the rampant curve B; then draw the diagonal ab ai, Ej and make b c perpen-
dicular to it, and equal to i c at ^; then draw the hypothenuse a c, and describe the angle-
rib Ej in the same manner as that of B.
To find the Length of the Jach-ribsy so that they shaUfit to the Bake of the Oroin.
Draw lines up £rom the plan to the arch, as at D, in the same manner as explained here-
tofore ; then the arch from a to a is the first jack-rib, irom, btob the second, and fix)m c to c
the third, &c.
To range the Angle^ibsfor these Oroins.
Get the ribs out in two halves, as in the last plate ; then the bottom of the ribs must be
bevelled agreeably to the ascent of the groin, and the plan of it must be drawn upon the
level, and from thence they may be drawn perpendicular from the plan to the rake of the
rib ; then take a mould to the form of the rib, or the rib itself, and slide this agreeably to
the rake to the distance that is marked upon the bottom to be backed ofi*; this will show
how much the rib is to bevel all around.
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GROINS. 85
PLATE 20.
QiveUy one of (he Bodp^rihe^ and ih^ WidOi tmd Height of (he Stde^x/rchy to find the Side and
Angle-rihaj ao (hoi (he Place of the Ihtereediony on the Plany ehaU be a straight Line.
Draw c e, at Fj perpendiculal* to the base of the body-rib B^ and equal to the height of
the side-arch 2>; from e, draw e a parallel to c &, and from a let fall the perpendicular abk;
draw k m, and k Z, which are the places on the plan of the intersection ; then^ the ribs D and
JS may be described from F^ as directed in the problem at Fig. 2, Plate 13 ; or they may be
obtained by intersecting lines, as at J. and O. The first method is, however, much the
easiest in practice ; for by using the points k, g, and fstE and Z>, as centres, the lines may
be struck by a chalk-line much sooner than the parallels at A can be drawn, and with greater
accuracy. But it must be recollected that four or five points will not be sufficient, in prao-
tice, for tracing the curve with accuracy ; and, therefore, a greater number must be found.
The way in which the groin is put together is shown below, in a manner sufficiently
intelligible for any workman. These groins are applicable when a window or door is to be
made in the side of a barrel-vault.
PLATE 21.
DEFINITION.
A Welsh groin is a groin whose side-arches are lees in height than the body-arch, a
both the side and body-archea are given semicircles, or similar segments.
Oiven, Oie Body and Side-rtba of a WeJsfi Oroin, to find a Mould/or Qie intersecting Rihs.
Divide half the side-rib B into any number of parts, and from these points 1, 2, 3, d, lej
fall perpendiculars to a h, its base, and produce them beyond ; also, from the name pointy
draw lines parallel to at, to intersect ef; transfer the divisions from e/to eg, and from th
divisions of e j draw lines parallel to p 7, the base of A, to intersect the body-rib A ^^
A, «, to,y; from these points, draw perpendiculars to pq, and continue them to intersect tin
perpendiculars from B, at the points /•, v,m,n; now trace a curve through thest^ points,
this will be the place of the intersecting rib on tlie plan ; then draw two other curve linef
on either side of, and parallel to this, to show the thickness of the rib on the plan ; on tl
inside curve draw chords from /, and on the outside curve draw tangents parallel to tha
chords. The distance between the chord and tangent shows the thickness of the stuff f
the intersecting rib. Draw perpendicular lines to the chords, and make the heights 1 1, 2
3 3, &c., at D, equal the corresponding heights at B. In a similar manner draw C; tha
will Cand I) be the moulds by which to cut the intersecting rib.
To i-ange iJie liiba so that they will stand perpendicular over the Plan.
At the points x, v, i, i, in the base of C, erect perpendiculars parallel to the ordinates of
and D, and make their corresponding Leiglits equal to the ordinates of B or- A; draw tbe
dotted curve h, », v, y, at C, and it will show how much is to be bevelled off that side of tfaa '
rib. In like manner, the otlier side Z* is to be bevelled. I
To find a Mould to bend under the intersecting Rib, ao that ilie Line of the intersecHnff An^d
may be marJced on the Rib according to the Plan. -^
Take the stretch-out of the under side of the rib D, by bending a thin slip of wood
around it; mark the different points, and transfer them to the straight line be, at E; then
make 1 n, 2 m, &c., at E, equal to 1 jj, 2 m, &c., on the plan. A curve drawn through
n, 771, 1, k, at E, will give the mould. The straight edge of the mould, when bent under,
nmst be made to correspond to the chord-line 1 2 3 c, at Z>; then, by drawing a line by the
curved edge of the mould, on the under side of the rib, we havathe true place of the ini
secting angle.
The method of framing the ribs is shown below, on the Plate.
itetJ
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GROINS. 87
PLATE 22.
7b describe the irUersecting or Angle-ribs of a Groin standing upon an octagon Pfan^ the Side
and Body-ribs being given both to the ea/me height.
Fig. 1. j^ is a given body-rib^ which may be either a semicircle or a semi-elb'pse, and
J. is a side-rib given of the same height ; 2> is a rib across the angles. Trace from E, the
base of both E and D being divided into a like number of equal parts, and divide the base
of the given rib A into the same number of parts. From these points, draw lines across the
groin to its centre at m, and from the divisions of the base of the rib D^ draw lines parallel
to the side of the groin. Then trace the angle-curves through the quadrilaterals, and the
result will give the place of the intersecting ribs. Draw the chords a b and b c, then mark
the moulds B and G from E or 2>, taking care not to mark them from the crooked line at
the base, but from the straight chords a b and b c.
lb describe and range the Angle-ribs of a Groin upon a circular Plan, (he Side and Body--
arches being given^ as in the last Groin.
Fig. 2. The ribs are described in the same manner as in the last example for the
octagon groin, or in the same manner as in Plate 21 ; and the ranging is found in the same
manner as is described in that Plate. E and F are the same moulds as are shown at B
and2>.
88 OBOINS.
PLATE 23.
Th& Slde-rib of a Oroin and Angles of Intersection being given straight upon a circular Plany
to fiiiA the AngU-riby and the Body-rib.
Fig. 1. Let the rib J. be supposed to be placed over the straight line a &, as its base,
which divide into any number of equal parts, as eight ; from the points of division draw lines
to the centre of the plan, to intersect the angles at a, 6, c, c?, c, /, g. From these points, erecl
perpendiculars to the line b dy and these being made respectively equal to those at J., will
give the curve of the rib Q. If from the points a, 6, c, &c., arcs be drawn with the centre of
the groin to intersect the base of G^ at 1, 2, 3, 4, 3, 2, 1, and perpendiculars be drawn and
made correspondingly equal to those of A, and O be traced to these points, then O will be
the body-rib.
■
To describe the Ribs of a Oroin over Stairs upon a circular Plany the Body-rib being given.
Fig. 2. Take the tread of as many steps as you please, suppose nine, from Ej and the
heights corresponding to them, which lay down at F\ draw the plan of the angles as in
the other groins, and take the stretch round the middle of the steps at Ey and lay it from
a to & at jPl Make d e perpendicular io dc at By equal io d e at F\ draw the hypothenuse
e c ; draw perpendiculars from dc up to jB, and mark B from Ay as the figures direct ; then
B is the mould to stand over ab at jPl Draw the chords a 4 and 4 m at the angles ; make
agy ihy perpendicular to them, each equal to half the height de, at B or F; draw the hypo-
thenuse g 4, and h m ; draw the perpendicular ordinates from the cords through the intersec
tion of the other lines that meet at the angles ; then trace the moulds D and O, from the
given rib A, and they will form the moulds for the angle or intersecting ribs.
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PLATE 24.
As oil the sectionB oF a sphere are circles, and those passing through its centre are equal, and
greateat which can be formed by cuttiog the Bpbere, it is evident that if the head of a niche is intended to
form a spherical surface, the best method is to make the plane of the back-ribs pass through the centre.
Hiia may be done in an infinite variety of positions ; but perhaps the best, and that which would be easiest
understood, ia to dispose them in vertical planes. If the head is a quarter of a sphere, the front-rib, and
the still-plate or springing, on which the back-ribs stand, will cnive equally with the vertical ones ; but if
otherwise, they will he portions of less circles. But it is evident, if the front and springing ribs are intended
to be arcs less than those of semicircles, either equal to each other or unequal, that as they are placed at
right-angles to each other, there ia only one sphere which can pass through them ; consequently, if the
places of the vertical ribs are marked on the plan, these ribs can have only one curve. In the former case
no diagram is necessary ; but in the latter it may be proper to show how the vertical ribs and their aituatioE,
on the front-rib are found.
To find the Hibsfor the Head of
t Spherical Niche, ike Elevation being a Semicircle, and the Plan
a Segment of a Circle.
\
From the centre G draw the ground-plan of the ribs as at A, and set out as many ribs upon the plan
as yon intend to have in the head of the niche, and draw them all out towards the centre at C. Place the
foot of your compass in the centre 0, and from the ends of each rib, at e and c, draw the small concentric
J dotted area round to the centre rib at m and n ; and draw m g and n t parallel to r 8, the face of the wall ;
- then from q round to o upon the plan, is the length and sweep of the centre-rib, to stand over a b ; and from
i round to o, the length and sweep of the rib that stands from c to d upon the plan ; and from g round to o
3 the sweep of the shortest rib, that stands from e to / upon the plan.
To bevel the Ends of the Back-riba against the Eront-rib.
The back-ribs are laid down distinct by themselves at C, D, and E, from the plan. Take 1 to 1',
A, and set it from 1 to 1' in i) ; it will give the bevel of the top of the rib JD. And from A, take fro
S to 2' upon the plan, and set &om 2 to 2' in the rib J^; it will give the bevel of the top.
To find the Places of the Baek-ribi where they are fixed upon the Front.
From the points a, c, and e, at the ends of the ribs, in the plan, at A, draw the dotted lines up to ti
front^rib, to o', c', and e', which will show where they are to he fixed upon the front-rib. The double c
upon the front-rib shows the ranging.
PLATE 25.
lb find Hie StU </ a S^aherioal Niche, ihe Plan and EJeoatian being given Segments of
Oirdee.
The elevation of the niche is shown at A, being the segment of a circle whose centre is t.
At .fi is the plan of the same width, which may be made to any depth, according to the
place it is intended for; its centre is c. On the plan B, lay out as many ribs aa it will
require, ana draw them all tending to the centre at c ; they will cut the plan of the front-rib in
g, /, e, d. ' Through the centre c, draw the line m n, parallel to a b, the plan of the front-rib ;
put the foot of your compass in the centre at e, draw the circular lines from a, g, /, e, a, to
the line m n, and make c s equal to ut; that is, make the distance from the middle of the
chord'line mnto a, the centre of the arch at O, equal to the distance from the middle of the
chord A. B at .^, to the centre at t; then place the foot of your compass in «, aa a centre, and
from the extremities m or n, describe the arch at C. With the same centre, draw another
line parallel to it, to any breadth that you intend your riba shall be ; then G is the true sweep
of all the back-ribs in the niche.
The points I, k, i, h, show what length of each rib will be sufficient from the point m ;
from A to Ri is the rib that will stand over dx; from t to m is the rib that will stand over
ejfi from Jb to mover/v; and from Z to m over j;' u ; the other half is the same.
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NICHES. 41
PLATE 26.
T7^ Plan of a Niche in a circular Wall being giveriy to Jmd ihe Frant-inh.
B is the plan given, which is a semicircle whose diameter is a by and a, i, kj ly m, hy the
firont of the circular wall. Suppose the semicircle j8, to be turned round its diameter a 6, so
that the point v may stand perpendicular over h in the fropt of the wall, the seat of the semi^
circle standing in this position upon the plan will be an ellipse. Then divide half the arch
of B upon the plan into any number of equal parts, as five ; draw the perpendiculars 1 dy
2eyB/y4tg. Upon the centre c with the radius c hy describe the quadrant of a smaller circle,
which divide into the same number of equal parts as are round B. Through the points 1, 2,:
3, 4, 5, draw parallel lines to a 5, to intersect the others at the points d, e,/, ^, h. Through
these points draw a curve : it will be an ellipse. Then take the stretch-out of the rib By
round 1, 2, 3, 4, 5, and lay the divisions from the centre both ways at Fy stretched out ; take
the same distances diycky fly g m, fix)m the plan, and at F make diyckyfly equal to them,
which will give a mould to bend under and mark the front-rib, so that the edge of the front-
rib will be perpendicular to a, ^, kylytn.
The curve of the front-rib at ul is a semicircle, the same as the ground-plan ; and the.
back-ribs sA C D and E are likewise of the same curvature.
The curve of the mould F will not be exactly true, as the distances diy ekyfly &c., at
By are rather too short for the same corresponding distances upon the mould at F\ but in
practice it will be sufficiently near for plaster-work; but those who would wish to see a
method more exact, may examine Plate 12, fig. 1, where C is the exact soffit that will bend
over its plan at B.
In applying the mould F when bent round the under edge of the front-rib, the straight
side of the mould F must keep close to the back-edge of the front-rib ; and the rib being
drawn by the other edge of the mould, will give its place over the plan.
6
4S NICHES.
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PLATE 27.
The Plan and Elevation of an ElUpaoiddl Niche being given^ to find the Ourm of the Ribs.
Fig. A. Describe every rib with a trammel, by taking the extent of each base from the
plan whereon the ribs stand to its centre, and the height of each rib to the height of the top
of the niche ; it will give the true sweep of each rib.
lb hack the Ribs of the Niche.
There will be no occasion for making any moulds for these ribs, but make the ribs them-
selves ; there will be two ribs of each kind. Take the small distances 1 e, 2 e?, from the plan
at By and put it at the bottom of the ribs D and Ey from d to 2, and 6 to 1 ; then the ranging
may be drawn off by the other 'corresponding rib ; or with the trammel, as for example, a.t
the rib Ey by moving the centre of the trammel towards e, upon the line e c, from the centre
c, equal to the distance 1 e, the trammel-rod remaining the same as when the inside of the
curve was struck.
Given one of the common Ribe of the Bracketing of a Oove, iofi/nd the Angle-iraehet for a
rectangular Room.
Let n be the common bracket, 5 c its base ; draw b a perpendicular to 5 c, and equal to
it. Draw the hypothenuse a Cy which will be the place of the mitre ; take any number of
ordinates in Hy {^rpendicular to 5 c, its base, and continue them to meet the mitre-line a c,
that is, the base of the bracket at J; draw the ordinates of I at right-angles to its base ; then
the bracket at ly being pricked from Hy as may be seen by the figures, will be the form of the
angle-rib required.
The way to obtain the angle-bracket of a common plaster cwnice is similar, and may be
seen at KL.
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PLATE 28.
ROOFS.
To find the pri/ifvcvp(d Lwvea of a Boof.
Let ABGDh^ the plan of a roof; take D L half the width of the roo^ and make D F
and C^j? equal to it. Join EF\ and having bisected EFeA ff, take (7 J7 equal to the height
of the roof, and perpendicular to EF^ and join OF and GF, and these lines will be the
length of the principal rafters. Join H C and HD ; produce HD, and take HI^ equal to
H O; join ICy and it will be the length of each hip-rafler. Draw a line, as a' l/^ perpendicu-
hxio HO; and with the centre &y draw an arc tangent to Oly cutting CHeA, df. Join df a!
and d! I/, and the angle a' d' V is what is generally termed the backing of the hip-rafter.
The bevel is shown at RySt being parallel \o G H.
To find the Bevels of a Purlme against a Htp-rafler.
Two cases are here exhibited ; the first is when the purline lies level, having two sides
horizontal. This is an easy case. The down bevel is shown at M, and the other at N. The
second case is that in which the purline stands at right-angles to the rafler. From the point
Cy the uppermost edge of the purline, as a centre, describe a circle. Draw the two lines b k
and e n tangent to the circle, and parallel to FD. Then fix)m the points g and A, where the
sides of the purline produced meet the circle, draw g % and h m also parallel to FD. Draw
h % and m n perpendicular to these lines, and join hf and fn. The down bevel is shown at
0, and the side bevel at P.
To find the Bevels of a Jach^rafler agaifist the Bip-rofkr.
The angle os/n is the side bevel, /a; being perpendicular to FD. The down bevel is
the angle F QH.
KOOVS.
PLATE 29.
Tbe lines for a roof baying a sqiiare or rectangular plan are finind bj the metbods
developed on Plate 28.
Fig. 1 i» tbe plan of a roof Since the plan is a trapezoid, tbe inclinations of tbe bips
will differ. Upon tbe lines AC and bd, as diameters, describe semicircles, catting the ridge-
line at K and r. From these points, draw lines to tbe extremities of tbe respective diameters,
and ihe»» lines will be the bases of right-angled triangles, whose height most equal that of
the rorjrf, and whose hypothenuse will equal in length tbe hip-rafters. Tbe bevels for backing
tbe bips are found as in Plate 28.
Fig. 2 is an octagonal plan for a roof, whose height is c b. Tbe lines are obtained as
in the preceding cases.
T/i^. Plan of a pdygcnal Roof being given, and a Sid&-rib of any Fomiy to find the Angle^ib
and the Form of the Covering.
Figs, 8 and 4. Let a b d be the given plan, and a 1 2 3 f be the given side-rib. Divide
a r \uU} any numlicr of parts, and, by means of lines drawn and measured as in tbe figure,
thmTi\)e tlie angle-rib b 1' 2' 3' o. Draw a e perpendicular to A b, and mark upon it the
divisions tjf ar. Make //', gg'j hh!, respectively, equal to dd\ ccf, hV, and through
/^ £^f ^f trace a curve ; then the figure e a b will be that part of the covering which is to
fio tx)nt over a b o on the plan.
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PLATE 30.
SKYLIGHTS.
One of the Ribs being given, the aquare Plan of the Opening or Welly and the octagon Curb ahovej
to find the Ribs and the springing Ourve, where the Foot of the Ribs comcy so thai the interior
Finish may correspond with the Curb,
Fig. 1. Let O be the given rib. From various points in it, draw lines perpendicular
to e &, and continue them around the curb, parallel to its sides. Erect the perpendiculars
to gf at Dy and make them respectively equal to those at C; then is D the angle-rih From
the points where the perpendiculars to eb intersect a b, erect others to a &, and make them
respectively equal to those at C; then will the curve s,t S he the springing curve.
The Plan of the Opening being sgnarcy to find ihe springing Curve amd (he Ribsy so that the
interior Finish may be spherical.
Fig. 2. Upon the side m n of the plan, describe the semicircle B ; this will be the
springing curve (for all sections of a hemisphere, perpendicular to its base, are semicircles).
From the centre of the circular curb at Ay draw the rib at Cy with a radius equal to half the
diagonal of the plan. All the ribs will have the same curve. Their several lengths may be
determined by drawing, with the centre of Ay small concentric arcs from the points a, &, c,
around to 1, 2, 3. Thence, draw perpendiculars \opSy meeting C at Vy 2\ 3'. The distances
from these points towards i will be the lengths of the ribs.
The vertioaX section of a segmental Dome passing through its Centrey ike Plan of ike square
Openvngy and the circular Curby being giveny to find the springing Curve and the Ribs
Fig. 3. Let the section over the diagonal ^ m, as shown in the figure, be given, and
find its centre s. Bisect mnhy a, perpendicular, and take a o, equal to s i. With the radius
o HI, draw the curve m b n^ and it will be the springing curve. The angle-rib is shown at C^
and the lesser ribs, all having the same curve, are obtained as explained in the last problem.
Fig. 4. These lines are obtained on the same principle as the others. The springing
curve is traced from C.
46 SKTLIGHTS
PLATE 81.
Bamng given ffie rectangular Plan of the Well and the elliptical Gwrh of a SJcylighty to find (he
Bibs and springing Curves, so that the Finish may he ellipsoidal.
Since all sections of a prolate-ellipsoid perpendicular to its axis are circles, the rib that
stands over 6 & at ^ is part of a circumference whose radius is the semi-conjugate diameter.
This rib is shown at D.
Since all sections of the ellipsoid not perpendicular to the axis are ellipses, all other ribs
will be parts of ellipses. The conjugate axis of these is in every case the same, and equals
the conjugate diameter of the ellipsoid. The semi-transverse axis in each case equals the
distance firom the centre s, through the desired rib, to the circumscribing ellipse, which is
described proportionate to the curb. The rib which stands over sp \b shown at E. The
distance cd at E equals that from p to the curb on the plan. The rib which stands over o s
is shown at ^.
Since all parallel sections of the ellipsoid are similar, the springing curves are similar to
the parallel section passing through the centre. Upon n 9, as a diameter, describe the semi-
circle at O; this is one springing curve. Taking the height of this as a semi-conjugate axis,
and m n as a transverse axis^ describe the semi-ellipse at B; this gives the other springing
curve.
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PLATE 32.
OF DOMES.
ToJiTid (he Covering of a hemispherical Borne.
Fig. 1. Describe a circle for the plan of the dome. Take the stretchout, a 6, of a quarter-
circrimference, b d, and place it from c to 4' at B. Take ah at B, of any length which the
board will admit, and describe upon it a eeniicirele. Divide half of this into any number of
equal parts, as four, and divide also c 4' into four equal parts, and through all the points of division
draw lines pitrallel to the base a 6. Make the lines 1' d, 2' e, 3'/, each equal to the corre-
Bponding lines within the semicircle, and trace the curve adef 4'. By this means, we obtain
the for^n of the boards, which, when bent over, will exactly cover the hemisphere.
7b find the Covering of a segmental Dome.
Fig. 1. Let c A be the height of the segment. Draw the cord g h, and the line 1 2 per-
pendicular to eg. Take the stretchout of the arc g h, and place it from c to 4' at C. Take
the line ab at C, equal to twice 1 g at A, and upon it describe a snlall segnaent of a circle
whose height shall equal 1 2 at A. Proceed as In the last case.
In a similar manner, the covering for various domical figures may be developed. The
method for obtaining the covering for an ogee figure is exhibited.
To find the Form ofUie Bom-da to cover a fiemisjyherical Dome hortzontallt/.
Fig. 2. Let the semicircle at A, whose centre is c, be a vertical section through tlie
centre of the dome. Divide the quarter-circumference d b into a number of equal parts, each
being equal to the width of a board. Extend the radius erf to m, and through the points
1 and 2 draw a line to t. Then, with ^ as a centre, describe arcs from the points 1 and 2,
which gives the form of the board. The others may be found in like manner.
Now, suppose tliat the centre of the board which lies between 5 and 6 is the last that
can conveniently be found — to find the others: with the centre y, continue the arc from 6
around to o, and through the points 5 and draw a line to meet the circumference at e. From e
draw radial lines to 5, 6, 7, S, &c. Also, draw lines from 6, 7, 8, &c., parallel to ai, meeting
c d at 6', 7', 8', &c.
Transfer 6'6 to a& at B, and take bd, equal to Q'fat A. Draw ad; make be equal to
a b ; draw d c, and bisect the angle dc h. At the middle point of the line a d erect fe per-
pendicular, meeting the bisecting Une at e. Through the points », c, and d, draw an arc.
If the line a d becomes too long to use conveniently, bisect also the angle e c a, as at G.
uid describe the arc a e upon the same principle.
48 ELLIPSOIDAL DOMES.
PLATE 33.
ELLIPSOIDAL DOMES.
Fig. A is the plan of a ellipsoidal dome ; B is the longest section, C the shortest 8ecti<m :
aamBy and hhxw Gy show how to square the purlines, so that one side may be fair with
the surface of the dome ; the dotted lines from aain B^ and hhin G, show how to get the
length and width of the purline in A ; but if the sides of the purline were made to stand
perpendicular over the plan, the curve of it would be found in the same manner ; then it
would require no more than half the stuff that the other would, and take only half the time.
Hoio to proportion the inside Gurh for the Skylight^ so that it will answer to the Surface
of the Dome.
Draw the diagonals i I and hh ^i A^ and let d e, or gf be the length ; then e g^ or df
will be the breadth of the curb ; because every section parallel to the base will be propor-
tional to the base.
The ribs (Figs. 4 and 5) in this case are obtained in the same manner as the ribs for an
ellipsoidal niche.
To find the Form of a Board to cover any Part of the Dome when bent up to the Grown.
Divide one-quarter of the base of the dome at D into three equal parts, rOjOp^pld. In
finding a board over sro in the plan, take the triangle sro ini), and lay it down at Fig. 2; then
draw the line s & a at right-angles to r o, and describe a rib, s 6 c, to the height of the dome,
and to the length of the perpendicular of the triangle sro. Divide it into five equal parts,
lay them along the line h a, and find the mould from the triangle sro, as the letters are
marked. The board (Fig. 2) will be found in the same manner.
Bemark. — In practice, you are to divide one-quarter of this dome into as many parts
as you think the breadth of the boards will require ; and the boards, when obtained by this
method, will fit with very great exactness. This is divided only into three, that the parts
may be clearly seen by learners. .
If the boards are obtained for one-quarter of the circuit, the corresponding boarjls
in the other three-quarters will not require other lines ; for every board in the first quarte
will be a mould for three more boards.
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[The designs which heretofore occupied this portion of the work have been laid
aside. They were, in their time, excellent ; but such advances have since been made in the
art as to render them useless to the carpenter of the present day. The designs substituted
are not offered as a complete series, to illustrate every case likely to arise, but are intended
to comprehend all the most approved principles of framing in use at this time. These prin-
ciples are difficult and important, and the designs which exhibit them will afford great
practical assistance to both architects and carpenters. — Editor.]
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PLATE 34.
SCARFING.
On this plate are exhibited ten different methods for scarfing timber. Some are secured
by iron pins, and in a few cases iron plates are also added. In cases where vertical pressure,
and not traction, is to be sustained, the best, but more expensive plan, is to bind the scarf
with iron straps and dispense with the bolts. Figs. 5 and 6 have no such aid, but are brought
tightly together by wedges. Many other designs for scarfing have been offei-ed, but being
more complicated, they are inferior to these simple forms.
(49)
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PLATE 35.
This ^ate cdnbiti ^fiuoBsd irrtiiirf ^^
^ven, that there maj be no A'fciH; h sadei^iBd
Tulee, which have rafereooe to the imi ctf'aakr h
In oak, ash, or elm, the whole In^ cf ife
thickness of the beam, when then are ao bolia or
In fir (or pine), the whole length of the xatfAaJd be aho^ tack* Hm
of the beam, when there are ik> bolu or stnpe.
In oak, aah, or elm, the whole length of a starf depCBfiap tm bolts oafy
three times the breadth of the beam ; and for fir beana, it AoaU be aboat ox tiam tbe
breadth.
When both bolts and indents are combined, the whole fength of tke acaif far oak and
hard woods may be twice the depth ; and for fir and soft woods, faor times Ae dt^iOi.
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PLATE 36.
TRUSSING.
Figs. 1 and 2 show the methods of ^ving the requisite stifiness to joists. Stout oak
laths, nailed on in the manner represented, improve their qualities in this respect very much.
Figs. 3 and 4 exhibit a method for trussing girders. A is the king-bolt^ and B is the
butt-bolt.
Fig. 5 shows a simpler and more effective method. The girder is hung on each side by
iron bars, secured by nuts at the ends. By turning these, any degree of camber may be
given to the girder.
Fig. 6 presents a still more effective method, though its use is limited to certain parts
of a building. The use of iron, in the manner here represented, is the great characteristic
feature in modem Carpentry. A girder, well constructed on this plan, will support a brick
wall built upon it.
62 TBUSSINa.
PLATE 37.
This Plate exhibits two designs for trussed partitions. It is a matter of great difficulty
to construct partitions so as not to spring and crack the plastering ; especially when ^there is
no support beneath, and when they are pierced by doors. In all such cases, they must be
trussed in a manner at least somewhat similar to those shown in the Plate. These will span
from twenty to forty feet.
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PLATE 38.
BOOFS.
Fig. 1 is calculated for a small building. The ends of the ridge-pole, purlins, and plates,
are shown in this and the following cases. The span may be twenty or twenty-five feet.
Fig. 2 is a roof for the purlins to be framed in. The dotted lines show the lower edge
of the common rafters. The span may be from twenty to thirty-five feet.
Fig. 3 is a much heavier roo^ the tie-beam being supported by two queen-posts, so as to
give room for a passage or apartment in the roof. It is calculated for a span of forty or fifty
feet.
64 BOOFS.
PLATE 39.
Fig. 1. This roof is somewhat similar to the last^ but essentially different in the con-
stPiction of the queen-posts. The span is forty or sixty feet.
Fig. 2 is a very light roof, having forty or sixty feet span. It may be used when the
ceiling below is to be arched. The sheathing may be laid directly on the purlins^ for a metal
covering.
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PLATE 40.
Fig. 1 is a very simple roof, designed upon the more recent principles which have revo-
lutionized Carpentry. An oaken block is placed between the upper ends of the principal
rafters, after the manner of a key-stone ; the grain of the block running in the same direc-
tion as that of the rafters. By this arrangement, the eflfect of shrinkage is rendered imper-
ceptible. An iron bar is passed through the block, and gives support and camber to the
tie-beam. The foot of each principal rafter is bolted with iron. The span of this roof may
be from twenty to thirty feet.
Fig. 2 is another roof, designed to contain an apartment. A horizontal piece is placed
between the bases of the queen-posts, that they may not be displaced by the lateral thrust
of the trusses. The span may be from forty to fifty feet.
l
56 ROOFS.
PLATE 41.
Fig. 1 is a design for a roof having very considerable span. It may extend from fifty
to eighty feet. We may here see the method of suspending the tie-beam by iron bars,
instead* of by king and queen-posts. Every purlin has its corresponding truss. 'Between all
pieces which do not come fairly end to end, oaken blocks are placed^ as before described.
The blocks are shaded, in this Plate. But what is here more particularly worthy of remark,
is the character given the design by the position of the trusses. It will be observed that
they have somewhat the appearance of lattice-work. This principle, we believe, is of Ame-
rican origin ; and is one of the most important improvements which have been introduced
into Carpentry of late years. The arches for domes and arched roofs, when trussed by this
lattice-work, are lighter, and more durable, than when constructed on any other plan. The
example here given exhibits the principle satisfactorily ; but it is capable of many impor-
tant modifications and applications, which should be thoroughly studied by our carpenters.
Fig. 2 is another roof, similar in principle, but heavier in proportion, and of shorter *
span. It may extend from fifty to sixty-five feet.
Where architectural effect is not an object, preference is now almost universally given
to flat roofs. The way in which this is managed in large roofs, is exhibited in both these
designs. The purlins, when necessary, are elevated on short, upright supports, standing on
tlie principal rafters ; and the walls are built considerably above the ends of the tie-beam.
By such an arrangement, any degree of weight may be given to the cornice.
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PLATE 42.
Fig. 1 is another design for a large, heavy roof, having a span of about thirty or sixty
feet. The oaken. blocks are, in this case, dispensed with. The ceiling-joists are shown
against the tie-beam.
Fig. 2 is a floor, designed to have a span of twenty or forty feet. The ends of both floor
and ceiling-joists are exhibited. The trussed girder which supports the floor, combines, as
may be seen in the figure, both the methods described in Plate 36, Figs. 5 and 6, with a
slight modification
Fig. 3 exhibits a section of the same floor, at right-angles to the above, through the line
a by Fig. 2. The shaded parts are the sections of the girders, and show their proper distance
^apart.
8
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PLATE 48.
Fig. 1 represent the framing of a church spire. It is quite light, but very strong. The
hoay n Mjuare, but the spire proper lias a conicnl finish. The whole rests upon frame-work
fupported by the outer walU; and in this ouse there need be no support immediately beneath
the «pir8.
Fig. 2 shows a horizontal section below the first window. Fig. 3 is a section throagfa
the second window. Figs. 4 and 6 are the naked flooring.
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PLATE 44.
Fig. 1 exhibits the framing for the roof of a church having a vaulted nave. It will be
observed that the opposite sides of this figure furnish two designs, either of which are suffi-
ciently substantial.
Fig. 2 is a Bimilar design, on a smaller scale. The roof, in this case, has a higher pitch,
but about the same span as the other.
60 ROOFS.
PLATE 45.
On this Plate jb shown a design for the roof of a small (jothic church, or hall. The
vault is ribbed, and the finish between the ribs is also exhibited. The ribs terminate in
some tracery on the upper portion of the walls, which is represented on the lower portion
of the Plate.
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PLATE 46.
This exhibits a design for an open-work Gothic roof for a church, hall, or open shed, for
large assemblies. The design is simple, and needs no remark. A different view of the
same design is shown on the bottom of thb Plate.
BOOFS.
PLATE 47.
We have here exhibited a design for an open-work roof for a small Gothic church or
hall, in the perpendicular style. A considerable degree of ornament always accompanies
this style. Below is shown the inner-wall decoration.
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ROOFS. 68
PLATE 48.
This Plate presents a design for an arched roof of very great span. It may extend as
much as one hundred and twenty feet, or even more, with safety, if the buttresses are very
strong. The arch is made by bending stout planks to the desired curve, and bolting them
together, which forms an arch of great strength and stability. Oftentimes, the arch is with
advantage made of the lattice-work before referred to. The stays which connect the arch
and principal rafter, consist of two pieces of plank, one on each side of the raft^er and arcb^
secured by long bolts.
64 DOMES.
PLATE 49.
DOMES.
Thb Plate exhibits a design for a dome. The framing is simple, although, from th»
nature of the sectional drawing, it appears complicated. A lantern is arranged for external
uffect, but does not extend to the interior.
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DOMES. 65
PLATE 50.
Fig. 1 shows the plan of the preceding design, taken from just above the oeiling-line
Fig. 2 is a horizontal section of the lantern, and Fig. 3 the roof of the same.
9
i \
06 BRIDQES.
PLATE 51.
On thiB Plate is shown the centring for an arch, as applied to a bridge. It is so con-
structed as to have no other support than at the abutments.
Fig. 1 is a vertical section, passing between the keynstone and the adjacent voissoir.
Fig. 2 is a side view of the cradling.
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WINDOW-CASING. 67
PLATE 52.
This Plate exhibits a design for the construction of a window-case, arranged for inside
shutters.
Fig. 1 is a horizontal section of the jamb. The shutter is folded into the jamb. The
circles above represent the weights with which the sash is hung.
Fig. 2 is a vertical section of the window-sill, having pannelling below.
I
JOINERY.
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HAND-RAILING.
PREFACE.
In that elegant branch of the building art called Joinery, Stairs and Hand-railing take
precedence. For the manner of finding the face and falling moulds, I have laid down correct
methods founded on the most obvious principles, and which have been put in practice by
myself, and by those who attended the instructions given by me, in this art, some years
since, and found to answer well in every case.
The superior advantages, in every respect, of the new plates on hand-railing, over those
published in the former editions of this work, will, I trust, be deemed a suflBcient reason for
the change made by me in this department of the Carpenter's Guide. I have retained Ihase
plates on hand-railing, by P. Nicholson, which are considered useful ; and hope that the
alterations made in this department of his work, will meet the approbation of Car])enter8
generally. In conclusion, I think it proper to say that, for the method of finding the butt
joints, Plate 55, I am indebted to an eminent stair-builder of this city, whose mechanical
skill in joinery, I, with others, hold in high estimation.
WILLIAM JOHNSTON,
Abchitect, Philadelphia.
PLATE 53.
Plan and Elevation of a Neicel PdsI Stair-case. Scale i an inch to a Foot.
To draw the Ramp.
Make A B equal to A (7, draw G D at right-angles with the rail, also produce le hori-
.zontal line E until it intersects at -D, which is the centre of the ramp.
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HAND-RAILING. 69
PLATE 54.
Plan of a stair-case, showing how to arrange the steps at the circle, so as to allow the
string-board to be formed without the easing (usually made) on its lower edge, caused by
making the tread of the steps greater or less at the circle than at the flyers ; this plan also
admits of the balusters being the same length, and nearly the same distance apart, as at the
flyers.
To draw the Plan.
Describe the equilateral triangle ABC; draw the right line E D, touching the face of
the string-board; produce GA and {75, to intersect it at D and E; then is the right line
ED equal to the semicircumference of the circle. Upon this line lay off* the tread of a step,
P, and draw lines from these points to the centre C; which gives the position of the risers
KK on the circular part of the string-board, at the points S S. The position of the same
risers on the carriage, is found by adding to the distance D P or Ey whatever may be
requisite to make a full tread, and place it from AtoW, wnich gives the position of the risern
NNand KK on the carriages.
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70 HAND-RAILING.
PLATE 55.
Fig. 1 is a plan of the rail for Plate 54 ; the stretch-out is found by the same method as
in the foregoing Plate.
7b draw the falling Moulds for the Bail.
Fig. 2. Let P be the pitch-board, and A the beginning of the circular part of the rail ;
from the line A jBT, place the stretxjh-out of the outside and inside of the rail, to G and D ;
and from the line D Bj set up the height D For G E^ which is equal to a rise and a half
Having found the stretch-out and height, place half the thickness of the rail on each side of
the central points Fj E and jBT, as denoted by the small circles ; then connect the upper
heights, E F, with the lower one, jBT, which completes the falling moulds.
To cut the centre or butt joints, divide the distance FE into two equal parts ; draw the
line EL, and at right-angles with this line, draw the lines F J bxA EK^ which gives the
joints required.
To find the Spring of the Plank.
Fig. 2. Take B Qy equal to (7 A, Fig. 1, and erect at (? a perpendicular to meet the
hypothenuse of the pitch-board produced at /; draw / parallel to 5 (?; then is the distance
O E the spring of the plank.
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nAND-RAILING.
PLATE 56.
7b draw the Face-mould /or the Rail.
Fig, 1. Draw the chord-line A D io the plan of the rail, also the base-line O P, oi any
convenient distance above it; make the length of this line equal to the dotted line on the
plan, and set the height NEof the falling moulds, Plate 55, from to .fl; draw HP; then
draw ordinates through the plan, perpendicular to the base-line, touching the chord R P;
also dra^v the ordinates BB, BB, &c., at right-angles with this chord, and make the distances
1 2, 3 4, &c., on the face-mould, equal to 1 2, 3 4, &c., on the plan.
The circular part being found, draw the straight part by keeping it parallel to the line
of the jomt S. In applying this mould to the plank, care is to be taken that the chord R P
be kept parallel to its edge.
Aa this method of finding the face-mould does not admit (if three inch plank be used)
of more than four or five inches of straight rail being attached to the circular part, another
method is shown below, by which the straight rail can be extended at pleasure.
Fig. 2. Let Q be the plan of the rail, and ABC the pitch-board ; draw the ordinates
1 2, 3 4, &c., at right-angles to GB; place the spring E, Plate 55, from Aio D; draw D E
at right-angles to A C. Make i> K aX N, equal to ED; draw EF, also GO, parallel to it;
draw the lines 0, 0, 0, &c., from the points made by the intersection of the ordinates with the
convex and concave sides of the rail, and produce them to the line G 0; draw C / at right-
angles with C A, make /J^ equal to EA, join JK, parallel to JK, draw the ordinates iX,
&c., and with Caa a centre, draw all the concentric lines to intersect with the line CH.
and continue those tines parallel to A G, and at the points of intersection with the oi-dinates,
trace the face-mould P, The dotted line S shows the over wood for cutting the i
butt joint.
he square or ^H
HAND-BAILINS.
PLATE S7.
Fig. 1 is the plan of a sturcase having eight windeis.
7b draw the falling Moidda.
F^. 2. Let AB\Xi the stretch-out of the convex side of the radl, and A I the stnugbt
portion attached to the circular part; draw the flyers ZTand (?, and the four winders ODE
and F\ also draw the base-line BI, at any convenient distance below the point 0\ raise the
rail five inches above the line of the noseings, from the point KiaJy also make the distance
A L equal to the distance A L, Plate 68, and proceed to draw the moulds as before described
in Plate 65.
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PLATE 58.
Jb draw the Face-movJd.
r
Let Am £ be tiie plan of the rail ; draw the chord A B, tiao the liiie OP^ panUlel to it;
makifr the lines EFG equal in height to the lined BL J^ Plate 57. Join QB; also draw U
parallel t6 <7D, and ^t the point J let fall the pe^pendicula]^ line JK Join K'iiy which
gives the directing ordinate on tiie plan; draw a sufficient number of ordinate 't6'ifie^t the
chord NO; also draw IP at right-angles with the line HO^ and with' the distance KMy
from the point /as a centre, bisect the line I Pat P. Jmn'^JPy which gives the directing
ordinate for the face-mould ; transfer the opdinates from the plan, and set them from the
chord N, anSl throughi the points trace the face-mould., Take the distance IS, set it fix)m
Tto ^on the chords and join 2im; with this bevel, the edge of the plaxik must correspon^f
before you apply the face-moulid to its upper and lower sui^ace.
. - J '
10
74 HAND-BAILINQ.
PLATE 59.
7b draw (he Skrcil of a Eand-Tuil.
Fig. 1. Make a circle 3} inches diameter; divide the diameter into three equal parts;
make the square in the centre equal one of these parts, and divide each of its sides into six
equal divisions, with the centres 1, 2, S, 4, &c. ; complete the outside revolution, set the
width of the rail from J^ to i^ and go the reverse way to complete the inside revolution.
If a scroll of less diameter be required, draw the dotted lines OA for the straight part
of the rail, and a scroll of one-quarter revolution less is given.
2b draw the falling MnUd/or the Scroll^
Ilg. 2. Take the pitch-board, and let J. J? be the stretch-out of the convex side of the
raiL Lay half the thickness of the rail on each side of the upper line of the pitch-board,
shown by the small circle; dni(w the line E Oy and the horizontal line FE, and complete the
under edge of the mould by intersecting lines.
In forming this easing, the mould should not come to a level at the joint but continue
to descend for a few inches past it^
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HAND-BAILINO. TA
PLATE 60.
7b dravo (he Face^mculd for (he Tuoist of (he ShroU.
Fig. 1. Draw a sufficient number of ordinates through the plan, and fix)m the points
1', 2% S% &c., draw ordinates at right-angles to the line 1' E^ and transfer the distances 1 1^
2 2, 3 3, &c., upon these lines, and trace the face-mould through the points.
The plumb-bevel is shown at D. jB ^ 1' is the pitch-board.
Ih draw (he Plan of (he curtail Riser.
Fig. 2. When the scroll is drawn, set the projection of the noseing without, and draw it
equally distant from it, which will give the form of the curtail step ; then set the distance
which the riser is back from the line of noseings, and you will get the form of the curtail
riser. This figure is drawn to a smaller scale.
HAND-BAILINO.
PLATE 61.
The methods for obtaining the moulda of hand-railing, exhibited in the two followinj
Plates, have never heretofore been published. They are founded on mathematical principle
and will be found to produce extremely accurate results. The lines are more simple,
much easier to draw, in practice, than those now in use; and the few principles here dew
loped may be applied to every ordinary case. — Ed.
Saving given the Man of (Jte Bailing, and tJie Tread and Ibeilion of Steps leading to t
Landing, to find the Face-mould.
Fig, 1. Let A D he the plan of the rail, either turning up towards I, or continuinj
towards H; draw the noseings in their proper positions ; lay down the distance D C ai ly C^
Fig. 2, and take the angle D CD', equal to the inclination of the railing, and draw D ly
perpendicular to CD'; take G A perpendicular \jo D C, and equal to £> G, Fig. 1. Then,
with CA as a semi-conjugate, and Z) C as a semi-tran averse axis, draw, with a trammel, t
quarter ellipse D A ; now, with B A half the breadth of the rail, draw a small circle,
W D B' at Z*' ; draw O O', &c., parallel io D ly ; and then, with C F ss & semi-tranavera
axis, and the same ee mi-conjugate, draw the inner edge of the mould, and it vrill be completfl
The plumb bevel is shown at K.
To find Ihefixlling Mould.
Fig, 3, Lay down a step at a 6 c, and take c D', equal to the stretch-out of the quadrant'
AD, Fig. 1; take bli, equal to the distance on the plan, from A (where the rail begins to
bend), to the riser of the top step, and ei-ect /* A at right-angles to « 6 ; take D ' i, equal to
six inches, or whatever distance }'ou wish to elevate the rail on tlie landing, and draw a 1:
through »', parallel to i>'c; also join n c, and continue it to meet the other line at rf;
B E and i D, each equal to the thickness of the rail, and draw Ef parallel to s rf ; now fro
A, trace any curve around to D (the arc of a circle is perhaps the best and easiest), and I
A' draw one parallel to this, and the falling mould is complete. The point A is appli
Aj Fig. 2, and the falling mould is bent around the piece, until the point D falls immedi
beneath D, Fig, 2.
To find the Thlchiess of the Stuff.
Fig. 3, Dtkw A'l parallel to c D', nnd take .A ' n equal to >4 C, Fig. 1 ; erect n/ f
pendicular to A' I, and from e let fall eg, perpendicular to/e; then will eg he the thicknei
of the stuff required to produce the rail.
Remark, — It will be observed that any length of straight rail may be formed in connej
tion with this piece; and by using sufficiently thick plank, the rail may be formed withot^
any joint whatever, except at D, Fig. 1. If a butt-joint is made at E, Fig, 3, the length c
the stuff required is equal to Efi
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PLATE 62.
lb find (he Face-mould for the Railmg of a SlairwaT/ having a semicircular Well, and StejM
of equal Tread througliout
Fig. 1. Tbe plan of the rail and stairs ia here exhibited. The steps do not wind, but
go up to a landing; and the rail continues an even ascent without any horizontal part, as in
the last case.
Fig. 2 exhibits the face-mould. Take Gb, equal Ui EF^ Fig. 1, the difference between
the radius G B, of the well, and the stretch-out of B O; make the angle Cba equal to the
pitch of the stairs, and draw a G at right-angles to CJ; extend a C, making Gc equal to the
radius B C oC the well ; join c and b; irom C let fall a perpendicular upon be, and extend it
beyond T; perpendicular to this, draw GE, and take Ca' equal to Ga^ through n', from 0,
draw O S, equal to B G. Fig. I, and at S erect the perpendicular S T; now, with O as a
centre, aud half the breadth of the rail as a radius, describe a small circle, and draw the
tangents m and n to the points where SO intersects the circumference; then take the dis-
tance T, and set it from G to F; extend the dividers from O to J^; then place one leg at
m, and mark m', and with one leg at n, mark n' ; around £ describe a circle to the breadth
of the rail ; now F G and E G, which equals B G, Fig. 1, arc trammel lines. Describe the
quarter ellipse F O E, and it will be the central line of the face-mould. Describe from m'
and n' the edges of the mould, tangent to the circle at E.
It now peraaina to find the butt-joints. Make Go equal to Oo', and through o draw
B y parallel to 6 c ; draw Be and Zi ' 6 at right-angles to b c, and through b' and G draw a line
cutting the mould in LD; this will be the upper butt. Also draw a line through C and B,
and from A draw downwards the outer edge of the mould parallel to GL, and make tlie inner
edge parallel to this; now, through L, draw a line (not represented on the Plate} parallel to
F, cutting tbe mould at K, and the lower butt joint may be made at JC, or any point above.
The mould is now complete. To use it, the points L and /f must be fixed to the edge of the
plank, and the --lould marked; then place the plumb bevel, shown at T, on the edge of the
plank at L, and it will give the point on the other side to which L must be affixed.
No faUing mould is requisite. A straight-edge bent amund the piece, when cut from
the board by the face-mould, gives the correct lines for dressing off the upper and lower sides
of the rail.
Remark. — This case is also applicable to a winding stairway. Instead of taking E F,
Fig. 1, equal to Gb, Fig. 2, take the sum of the heights of the risers between B and O, Fig.
1, and having added five inches, lay down this length from A to B, Fig. 3; then make the
angle A GB equal to the inclination of the rail {A B G being a nght-angle), and take B E,
equal to B G, Fig. 1, and erect EF perpendicular to B G; now use the right-angle triangle
EF G, Fig. 3, in the place of Ga b, Fig. 2, and it will give the face-mould for the rail of the
winding stairs. In this case, however, falling moulds will be necessary. They may be found
in the vny indicated upon Plate 57.
T8 HAND-BAILING.
PLATE 68.
lb draw (he Form of a Hand-rail.
Fig. 1. Make an equilateral triangle, vwt, upon the width of the rail, and divide it
into five equal parts, and firom one part on each side draw z s and y u ; then t, g, and m, are
the centres, I m being made equal tolg. The centres are found the same for the other side.
The Form of a Bail heing given, to draw (he Mitre^oap. •
Fig. 1. Let the projection of the cap be three inches and a half, and make the distance
of the inside circle from the outside circle, equal to the projection of the nose on each side
of the rail, and draw tbie mitre n o and n' o ; then continue parallel lines down to the mitre
n' o, put the foot of your compass in the centre of the cap, and circle the parallel lines round
to a', c', 6^, ^, and i'y and draw the ordinates a' V, cf d', ef /', &c., and then mark out the
cap for the rail according to the letters.
7& draw the Form of (he Gap; (he MUre to come to (he OenJtre.
Fig. 2. In every case draw the lines which are perpendicular to the diameter o
rail, to the mitre ; and then, with the apex of the mitre, as a', for a centre, describe the arcs
around to the diameter of the cap ; then make a' V, d d\ ef f, &c., respectively, equal to a &,
ed^ef &c., and it will fffY% the form of the cap.
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PLATE 64.
7b draw the Soroll o/ a Eand^raH.
Fig. 1. Make a circle with the centre c, three inches and a half in diameter, and divide
the diaraeter into three equal parts ; make a square about the centre of the circle, whose
aides are equal to one of these parts, and divide each of its sides into six equal parts. The
square must then be divided by lines, as shown, full size, at Fig. 2. Now, with the points
1, 2, 3, 4, 5, 6, as successive centres, describe first the arc from i to k, Fig. 1, then from k to
h', from i/ to m, and so on around to a. This will give the outside spiral. Set the width of
the rail from a to/, and go the reverse way for the inside spiral. This will complete the
scroll.
To draw (he curtail Step.
Fig. 1. Set the ballusters in their proper places on each quarter of the scroll ; the first
balluster shows the return of the noseing round the step, the second balluster is placed at the
beginning of the twist, and the third balluster a quarter distant, and straight with the front
of the last riserj then set the projection of your noseing without, and draw it all round
equally distant from the scroll, which will give the form of the curtail step.
Tofiiid the falling Mould.
Fig. 3. At a & c is the pitch-board ; the height is divided into tax parta, to ^ve the
level of the scroll ; the distance ad \9 from the face of the riser to the be^nning of the
twist ; and the distance from rf to ft is the stretch-out from a, the beginning of the twist
round to h, Fig. 1 ; divide the level of the scroll, and the rake of the pitch-board, into a like
number of parts, and complete the top edge of the mould by interBCcting lines, and the under
edge parallel to it to the depth of the rail.
The outside falling mould, Fig, 4, is found in like manner.
The method for obtaining the thickness of the stufi* requisite for cutting the scroll is
also shown at Fig. 3.
HAND-BAILING.
PLATE 65.
As the method of getting a ecroll out of a solid piece of wood, having the grain of the
wood to run in the same direction with the rail, is far preferable to any of the other methods
with joints in them, being much stronger than any other scroll with one or two joints, and
much more beautiful when executed, as no joint can be seen, and consequently no difference
in the grain of the wood at the same place ; I shall here give a specimen, the method for
describing a scroll being already given in the last Plate; and Ukewise the falling mould.
How to fitid Oie JFace-mould.
Fig. 1. Place your pitch-board, a'J/&,aa in the last Plate ; then draw ordinatea across
the Bcroll at discretion, and take the length of the line d b', with its divisions on the longest
side of the pitch-board, and lay it on d' h' in Fig. 2 ; then, the perpendiculars being drawn, it
may be traced from Fig. 1, as the letters direct.
How tofitid the Thicjcness of the Stuff.
Fig. 3, Let a 6 c' be the pitch-board, and let the level of the scroll rise one-sixth, aa in
the last Plate ; and from the end of the pitch-board, at b, set from b to d half the thickness
of the balluster, to the inside; then set from d to e half the width of the rail, and draw the
form of the rail on the end at e, the point h being where the front of the riser comes, then
the point e will be the projection of the rail before it; then draw a dotted line to touch the
nose of the scroll, parallel with & b, the longest side of the pitch-board ; then will the distance
between m and n be the thickness required. Much lighter stuff may be used, by gluing a
piece on the under side, vs shown at m.
Figs. 4, 5, and G, show the plan, face-mould, and thickness, of another scroll, of a difffr
rent pattern. The principles of the drawing are the same as in the other case.
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HAND-BAILING.
PLATE 65.
As the method of getting a ecroU out of a solid piece of wood, having the grain of the
wood to run in the same direction with the rail, is far preferable to any of the other methods
with joints in them, being much stronger than any other scroll with one or two joints, and
much more beautiful when executed, as no joint can be seen, and consecpiently no difference
in the grain of the wood at the same place ; I shall here give a specimen, the method for
describing a scroll being already given in the last Plate ; and likewise the falling mould.
How lofiiid Ote Face-mould.
Fig, 1. Place your pitch-board, a' ^ &, aa in the last Plate ; then draw ordinates across
the scroll at discretion, and take the length of the line db', with its divisions on the longest
side of the pitch-board, and lay it on d' f/ in Fig. 2 ; then, the perpendiculars being drawn, it
may be traced from Fig. 1, as the letters direct.
ffoio tofiiid (lie Thickness of the Stuff.
Fig. 3. Let o J c* be the pitch-board, and let the level of the scroll rise one-sixth, as in
the last Plate; and from the end of the pitch-board, at h, set from 6 to d half the thickness
of the balluster, to the inside ; then set from d to e half the width of the rail, and draw the
form of the rail on the end at e, the point b being where the front of the riser comes, then
the ix)int e will be the projection of the rail before it; then draw a dotted line to touch the
nose of the scroll, parallel with & b, the longest side of the pitch-board; then will the distance
between m and n be the thickness required. Much lighter stuff may be used, by gluing a
piece on the under side, as shown at m.
Figs. 4, 5, and G, show the plan, face-mould, and thickness, of another scroll, of a diffe-
rent pattern. The principles of the drawing are the same as in the other case.
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PLATE 67.
To show the proper Ttviet af a Bait and Tktdatem of Stt^^ tmi ihe Faea-moi^ds.
] ;. 1. Let A be the plan of a rail, aod Fig. 2 the falling mould, as completed. Let
"' nf the rail be so placed that the chord A o be parallel to the base v o of the falling
.. . 2, and let B b' be the separation of the straight and circular parts of the rail, O B,
y beir the quadrantal part of the rail, and b' a', a b the etraigfat part ; divide the concave
side B CDo, of the quadrautai part, into equal parts at the points B, c. d. and o; through all
the points B, c, d, &c., draw lines at right^angles to the chord A o, cutting it in 1, 2, 3, &c.,
and produce them upwards to the points/*/ h, &c. Let M B N, Fig. 2, be the section of a step,
and upon M o, make B A equal to b a. Fig. 1. Extend the parts A B, b c, c D, &c., upon the
base M 0, Fig. 2, from A to b, from B to c, from c to D, &c. ; from the points a, b, c, d, &c.,
draw lines perpendicular to M 0, cutting the lower edge of the falling mould at E, p, g, h, &c.,
and the upper edge at i, J, k, l, &c. In Fig. 1, draw any line, abed, &c., parallel to the
chordAO; make ac, 6/, c J, d/i, &c., respectively equal to AE, BF, CG, DH, &c., Fig. 2; also
in Fig. 1, make a i, bj, c k,dl, kc, equal to a i, b j, c k, d l, 4c., Fig. 2. Through the
points e, /, g, h, &c., FJg. 1, draw a curve ; also through the points i, j, k, J, &c., draw another
curve, and these two curves will complete the projection of the falling mould. From the
point s, Fig. 1, radiate the lines c c', d d', &c., cutting the convex side at c* r/, &c. ; from the
points a', b', c', d', &c., draw the lines A' i', B'f, c' ly, d' h', &c. ; draw e ^, i i', jj', k k, h h',
&.C., parallel to the chord a o. Through the points »', /, ^■', &c., draw a curve until it inter-
sects with the curve i,j, k, &c. ; this will form the projection of the top of the rail piece. In
like manner the under parts which appear a.tgg', hh', will be completed in the same manner,,
so that b is the whole appearance of the solid, qr* rq' and ii' etf being sections, or the ends
which join the contiguous parts of the rail.
To trace the face-mould at C; join i' r', and let the perpendiculars cut ir at 1, 2, 3, &c.
Draw the ordinates \hu,^cv,$dw, &c., perpendicular to i' r* ; make 1 6, 2 c, 3 rf, &c., equal
to I B, 2 c, 3 D, &c., at A ; also, at G, make 1 u, 2 «, 3 w, &c., equal to 1 u, &c. ; also make x o
equal to xo', ia' equal to I a'; draw by parallel to ta', and ay parallel to tb, and ta' yb
will complete the straight part of the face-mould: join r' o; draw the concave curve
bedr, and the convex curve yuvwo, which completes the curve part, and the whole of the
fttce-mould.
Fig. 3 shows the projection and face-mould for the quadrantal part of a rail. The
principles of projection, and manner of drawing the face-mould, is the same as wha£ has now
Jiijun Hhown. The figure is introduced to show how much less wood the part of the rail
niqilircM fi-otii a quadrantal plan, than when a straight part of the rail is taken in; and the
llltmi of the Htraight rail tliat is taken in, the greater will be the deflection from the chord;
HiiiH, III KiK- 'li the distance between the chord/ r, at any point to the nearest point of the
|irii|t'rllon, in nmi^li less than the^distance, Fig. 1, from a corresponding point in i'r to the
iX'iit'KHl piiiiit of the projection.
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HAND-RAILING.
PLATE 65.
As the method of getting a scroll out of a solid piece of wood, having the grain of the
wood to run in the same direction with the rail, is far preferable to any of the other methods
with joints in them, being much stronger than any other scroll with one or two joints, and
much more beautiful when executed, as no joint can be seen, and consequently no difference
in the grain of the wood at the same place ; I shall here give a specimen, the method for
deacribing a scroll being already given in the last Plate; and likewise the falling mould.
How to find Uie Face-mould.
Fig. 1. Place your pitch-board, a' 1/ c^, as in the last Plate ; then draw ordinates across
the scroll at discretion, and take the length of the line db', with its divisions on the longest
side of the pitch-board, and lay it on d' i/ in Fig. 2 ; then, the perpendiculars being drawn, it
may be traced from Fig. 1, as the letters direct.
How to find five Thickness of the Stuff.
Fig. 3. Let a J c' be the pitch-board, and let the level of the scroll rise one-sixth, as in
the last Plate; and from the end of the pitch-board, at h, set from h to d half the thickness
of the balluster, to the inside ; then set from dio e half the width of the rail, and draw the
form of the rail on the end at e, the point h being where the front of the riser comes, then
the point e will be the projection of the rail before it; then draw a dotted line to touch the
nose of the scroll, parallel with & b, the longest side of the pitch-board; then will the distance
between m and n be the thickness required. Much lighter stuff may be used, by gluing a
piece on the under side, as shown at m.
Figs. 4, 5, and 6, show the plan, face-mould, and thickness, of another scroll, of a diffe-
rent pattern. The principles of the drawing are the same as in the other case.
HAND. RAILING.
PLATE 6i
Ibjind Uie Moulds /or ex&mtiiig a Rail with a Semicircle of Windere,
Fig. 1. The falling mould as here drawn does not follow the line of noseinga, but ib
raised six inches, as is the practice with several hand-railere, in order that the mil should not
approach nearer to the noseinga of the winders than to the flyers. This example is adapted
to a stair with ten winders in the semicircumfereuce.
Fig. 2. The plan and face-mould of the lower quarter winders.
Fig. 3. The plan and fiice-mould of the upper quarter winders. A b c the convex side
of the quadrantal part of the plan, c d a straight part intended to be wrought on the sanje
piece with the twisted part. In this method, the plane of the top of the face-mould is sup-
posed to rest upon the upper extremities of three straight lines or slender rods perpendicular
to the plane of the seat of the said face, and these three perpendicular lines to rise from three
points, in a line dividing the breadth of the plan everywhere into two equal parts of the rail
piece ; and these three points to be so situated that one may be at each extremity, one at the
end of the quadrant, and one at the end of the straight piece, and the intermediate point at
the intersection of a pei-pendicular, drawn from the centre of the plan of the rail piece to a
straight line joining the two extreme points. Let each of the upper extremities of the three
perpendicular lines be called resting-points, and let the feet of the perpendiculars be called
the foot of the heights of the rail piece, which will therefore be the same as the seats of the
resting-points; and let the three perpendicular lines themselves be called the heights of the
rail piece, and their places distinguished by the lower height, the middle height, and the
upper height.
Let a, Fig. 2, be the foot of the upper height, h the foot of the middle height, and d the
foot of the lower height ; join a d ; draw af,l/g, and d h, perpendicular to ad; the middle
one, bg, being drawn from the centre E, Let a, b, c, d, Fig. 1, be the points corresponding
to A, B, c, D, in the convex side of the plan. Fig. 2, and let a f, b g, and d h, be the heights
of the rail piece, Fig. 1 ; make a/, f g,dh, Fig. 2, respectively equal to A p, b G, d h, Fig, 1 :
join /A, Fig, 2 : draw g i parallel to bd, cutting /ft at i; draw i k parallel to gb, cutting ft d
at k: draw g Im perpendicular to/h, cutting /A at /; join bk; from i, with the distance 6 A,
describe an arc at m ; join i m : draw d' p parallel to d a from the extremity d of the concave
side; produce the convex quadrant c b A to Q, and the concave quadrant c' b' A to p, and
riidiate the line p Q from the centre e, which will complete the whole plan of the rail piece;
the part p a' a q will make a suflBcient allowance for the cutting of the joint. Draw ordinateit
parallel to ft i, cutting the chord i^ p, the concave side of the plan, and the convex side of the
same : produce ft A; to meet D' p in t, and produce mi to u, making i u equal to kt: through
PUBLIC UBRAR^
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HAND. RAILING. M
PLATE 69.
Jb find the Face-mould far (he BaU of a Stair-oaae hiving a Landing, ao (haij when aet to its
proper Bake, it wUl stand direcUy over (he givers Plan.
Fig. 1. Draw the central line h q parallel to the sides of the rail ; on the right line h q
apply the pitch-board qca\ from qXoa, draw ordinates nd',meylfyhg, and ih, at discre-
tion, taking care that one of the lines, as h g, touch the inside of the rail at the point g, so
that you may obtain the same point exactly in the face-mould ; then take the parts qu^uvy
vWy wxy xy, and apply them at Fig. 2, from qu, u% vw, wx, xy; from these points draw
perpendiculars to ^ ^, and mark their lengths from the plan, Fig. 1 ; then Fig. 2 will be the
mould required.
Ih find the faUing Moiild.
Divide the radius of the circle, Fig. 1, into four equal parts, and set three of these parts
from 4 to 6 ; through n and v, the extremities of the diameter of the rail, draw b n and b v,
to cut the tangent line at the points o and d ; then will odhe the circumference of the rail,
which is applied from c to d, at Fig. 3, as a base-line : make c e the height of a step ; draw
the hypothenuse ed; si the points e and c2, apply the pitch-board of a common step at each
end of their bases, parallel to cd; make c2/ equal todg, i{ it will admit of it, and by these
lengths curve o£f the angles by the common method of intersecting lines ; then draw a line
parallel to it, for the upper edge of the mould.
Plate 70
8& HAND- RAILING;
PLATE 70.
lb draw a/aUmg Mould for a Rail having Winders all round the circular Ihrt; thence to find
ike Face-mould.
To describe every particular in this, would almost be repeating what has been already
described. The heights are marked the same upon the falling mould, Fig. 2, as they are at
the face-mould, which will give the heights of the sections of the rail ; and the face-mould.
Fig. 1, is traced from the plan according to the letters. Fig. 3 shows the application of the
mould to the plank; take the bevel at h, Fig. 1, and apply it to the edge of the plank at
Fig. 3, and draw the line h c ; then apply your mould to the top of the plank, keeping one
comer of it to the point J, and the other corner close to the same edge of the plank ; then
draw the top face of the plank by your mould ; then take your niould, and apply it to the
under side at c, in the same manner.
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88 HAKD-RAILING.
PLATE 72.
Plan and Elevation of an eUipticcU Stair-caae.
In every kind of stair-case whatever, the breadths of the heads of the steps are always
reckoned on a line, bisecting their length, or at 18 inches distapt from the rail. In this
example, the steps are divided into equal parts, both at the rail and at the wall. This divi-
sion will make the falling mould straight on the edges, and consequently will form an easy
skirting as well as an easy rail.
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PLATE 78.
To draw the Fac&moulds for the Bailing of an eUiptio Stair-case.
The plan and section being laid down as in the preceding plates, it will be observed that
the ends of the steps are equally divided at each end; that is, they are equally divided round
the elliptic wall, and also at the rail. In this Plate, the rail is laid down to a larger size than
that m the last Plate; the plan of this rail must be divided round, into as many equal parts
as there are steps ; then take the treads of as many steps as you please, suppose 8, and let
hh, at Fig, 1, be the tread of eight steps; on the perpendicular h m set up the height of as
many steps, that is, 8 ; and draw the hypothenuse m h, which will give the under edge of
the falling mould. The reader will observe that this falling mould will be a straight line,
excepting a little turn at the landing and at the scroll, where the rail must have a little bend
at these places, in order to bring it level to the landing, and to the scroll ; then mark the
plan of your rail in as many places as you would have pieces in your rail (in this plan are
three) ; then draw a chord line for each piece to the joints; also draw lines parallel to the
chords, to touch the convex side of the plan of the rail; from every joint draw perpendicu-
lars to their respective chords. Now the tread of the middle piece at c being just 8 steps,
the height of the section from /t to m is 8 steps; and the section m n is the same as m n on
the falling mould, and the section h i is the same height as h i upon the falling mould ; draw
a line to touch the sections, and complete your face-mould aa in the foregoing plates: in each
of the other pieces at E and G, tbe number of treads being 6 ; therefore, from your falling
mould set the stretch of six steps, from h to h'. Draw h' I parallel to A n, then k'kl will
give the height of the sections at d and E : everything else agreeably to the lettei-s.
The stretch-out of 8 steps, or any other number, is not reckoned on the chord; but it is
the stretch-out round the convex side of the rail, or what most people call tbe inside.
HAND-RAILING.
PLATE 74.
Haw to diminish the Shaft of a Column by the ancient Method.
Fig. 1. Describe a semicircle at the bottom; let a line be drawn througb the diameter ]
at the top, parallel with the asiB of the column, till it intersects the eemicircle at 1, at thai
bottom; then 1, 1 at the bottom will be equal to 1, 1 at the top; divide the arch into fourJ
equal parts, and through these points draw lines parallel to the base, the height of the column
being also divided into the same number of parts, and lines drawn parallel to the base, then
the column is to be traced from the semicircle, according to the figures.
How to describe the Column hy another MclJiod.
Fig. 2. Take the semi-diameter a h at bottom, and set the foot of your compaps in the
top at c, and cross the axis at 8, and draw the line a 8 on the outside, parallel to fc 8 on th* J
axis, and divide each of these lines into eight equal parts, and set the diameter a h at the |
bottom, along the slant lines 1 1, 2 2, 3 3, &c.j from the axis; this will give the diminishing J
of the column.
Bern to diminish ilie Column by Lines drmon from a Centre at a Distance.
Fig. 3. Take the serai-diameter a J at the bottom, set the foot of your compafls in c at
the top, and cross the axis in the point (/; continue ct/ at the top, and a t at the bottom, to
meet at e; then draw from e as many lines across the column as you please, and take the
diameter a ?» at the bottom, and prick each line upon the axis equal to b a, which will give
the swtll of the column.
To diminish a Column by LalJis, upon i/ie same Principle.
Fig. 4. The point e being foimd, as in Fig. 3, groove a rod d b, and lay the groove upon
the axis of the column, and groove the describing rod upon the under side, and lay the groove
upon a pui (ixed at e; then fix a pin at y, to run in the groove upon the axis of the cohimn,J
and the distance of the pencil at/, equal to 6 o, then move the pencil at/, it will describe the I
curve.
How to make a diminishiog Mule.
Kg. 5. Divide the height of your rule into any number of equal parts, as 6 ; draw linesl
at right-angles from these points across the rule, and divide the projection of the rule at the i
top, that is, half of what the column diminishes, into the same number of equal parts. Put
a pin or brad-awl at a; lay a ruler from a to 5, mark the cross line at /; then lay a ruler
from 4 to a, and mark the next cross line at e; then lay tjie ruler from 3 to a, mark the next
at d, and so on to the bottom ; bend a slip round these points, and draw the curve by it, and
it will give a proper cui-ve for the side of the column.
This is the readiest method, and gives the best curve of any that I have tried.
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SASn-WORK.
PLATE 79.
The Plan and Eleoatum of a circular Sash, iii a circular Wall, being given, to find Ote Mould
for tlie radial Bars, so tliat they shall be perpendicular to tJie Plan.
Fig. 1. Draw perpendiculars from the points 1, 1, 1, 1, &c., at A and b, in the radial
bars, either equally divided, or taken at discretion, down upon the plan to 1, 2, 3, 4, 5, 6, 7,
at c and d ; and draw a line from tlie first division upon the convex side parallel to the base;
then draw ordinatcs from 1, 1, 1, 1, &c., at right^angles to the radial bars, at A and B, which
being pricked from the planB at d and o, will give a mould for each bar; and the bevels upon
the end will show the application of the moulds.
7b JtnJ the Veneer of the ArcJi-bar. ■
Fig. 2. To avoid confusion, I have laid down the plan aijd elevation for the head of the
sash below. The stretch-out of the veneer is got round, 1, 2, 3, 4, 5, 6, on the arch-bar.
which, being pricked from the small distance on the plan at m, will give the veneer above, at e.
To find Ote Face-moidd for the &tHh-}iead.
Fig. 2. Divide the sa-sh-head round into any number of equal parts, at G, and draw
thera perpendicular to the base at h ; draw the chord of one-half of tiie plan at h, and draw
a line parallel to it to touch the plan upon the back side; then the distance between these
lines at h, will show what thickness of stuff the head is to be made out of; and from the
intersecting pointa on the back side, draw perpendiculars from the base of the face-mould,
which, being pricked from tlie elevation, as the figures direct, will give the face-mould.
To find the Moulds for giving the Fbrm of the Head, perpendicular to (he Plan.
Fig. 3. The baae of l is got round the arch 1 2 3 4 5 6, at f. Fig. 2, and the base of k
is got round ahcdef g, also at f, and the heights of the ordinates of each are pricked either
from H or I, which will give both moulds.
Note. — The face-mould nt a. Fig. 2, must be applied in the same manoer as in groina; ao that the ensb-hcod mast
be bevelled by sbifting tbe mould o, on each side, before you cao apply tbo moulds k and l, Fig. 3 ; the biuck lines at
K aad L arc pricked from the plan, al h; these black lines will exactly coincide with the front of tho rib when bent
round; b line being drawn by the other edge of the moulds, will be perpendicular over its plau, aud the tbiokneBS of
the sash-frame towards the inside will ba found Dear enough by guaging from the outside.
92 ARCHITBAYK.
PLATE 76.
Fig. 1. The plan and elevation of an architrave or archivolt for a circular window^ in
a circular wall.
• To find the Form of the Veneer.
Fig. 2. Lay down the line 6^ 6^^ equal to the douhle stretch-out of the line ahcd, &c.,
Fig. 1 ; erect perpendiculars from the corresponding points^ and by measurement we may
obtain the arch 5 4 3 2^ &c., which is the form of the veneer.
The first veneer is to be partly cut out of the solid architrave as far as % 3 and k 3^ on
each side^ to join to the middle piece that lays between these joints; break the joint with the
next veneer that is between p & and m n, or 6^ s, &c.
I'i.it.' 71"
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MOULDINGS. ».«
PLATE 77.
2b deacrUe the Angle Bars for Shop Fronts.
Fig. 1. B is a common bar, and c is the angle bar of the same thickaess ; take the
raking projection 1, 1, in c, and set the foot of your compass in 1 at B, and cross the middle
of the base at the other 1 ; then draw the lines 2, 2 ; 3, 3, &c., parallel to 1, 1 ; then prick
yonr bar at fix)m the ordinates so drawn at b, which being traced will give the angle bar.
lb draw the Mitre Angle of a Commode Frardfor a Shop.
Fig. 2. Divide the projection each way in a like number of equal parts; then the
parallel lines continued each way will give the mitre.
*
To find the Rdkmg Mouldings of a Pediment.
Fig. 4. Let the simarecta on the under side be the given mouldings and let Imes be
drawn upon the rake at discretion ; but if you please, let them be equally divided upon the
simarecta, and drawn parallel to the rake ; then the mould at the middle being pricked off
from these level lines at the bottom, will give the form of the face. The return moulding at
the top must be pricked upon the rake, according to the letters.
The cavetto, Fig. 3, is drawn in the same manner.
N. B. If the middle moulding. Fig. 4, be given, perpendiculars must be drawn to the
top of it ; then horizontal lines must be drawn over the mouldings at each end, with the
same divisions as are over the mouldings ; and lines being drawn perpendicularly down, as
above, will show how to trace the end mouldings.
94 MOULDINGS.
PLATE 78.
RaMng MouTdingSy and Mouldings of different PrqjectUmB.
Figures 1 and 2 show how to trace base mouldings for skirting to stairs, upon the same
principles as shown in the last Plate; at the bottom are given two methods of mitring
mouldings of different projections together.
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MOULDINGS. 95
PLATE 79.
Qiven the Form of a CoTrdce^ to draw Utoa greaier Proportion.
Fig. 1. Let the given height of the cornice he ad\ set one foot of your compass in a,
and cross the under side at h with that height, and from the point o draw the line o c2 at
right-angles to a & ; then the height of all your mouldings will be the parts of a h^ and the
projections the parts of c c? in proportion.
Note. — a /shows another height; c e its projection in proportion to that height
Th diminish a Cornice in the Proportion of a greater.
Fig. 2. Describe equilateral triangles on the base and projection, as at A, and make if
and ig equal to the intended height, and draw the line/^ across the triangle, which will
give the heights in proportion to a 6 ; put the foot of your compass in i as a centre, and
circle he round hhj and draw the dotted line At, cutting fgiiik\ then set oS ie and idj
each equal to ^ A; ; draw d e ; then take the divisions of e dj and set them from fiom\ in
the same order draw perpendiculars : it will give the diminished cornice at A.
To do this by another method, as shown at B, let the given height be a 5, and draw the
hypothenuse ag, and lines being squared up io-ah^ from the divisions of ag^ will give the
heights ; and if you draw the line ^ c2 at a right-angle with a g, then d o will give the projec-
tion in proportion, when returning upon d c.
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96 MOULDINGS.
PLATE 80.
MOULDINGS UPON THE SPRING.
To find the Sweep of a Mmddmg to he bent upon the Spring rou/nd a circular Ck/linder.
In Fig. 1, which stands upon a semicircular plan, make a c equal to the height of your
moulding, and make a b equal to the projection ; describe the form of the moulding, and draw
a dotted line to touch the face of it ; then draw the line ed to meet in the centre of the body
at (2, so as to keep your moulding to a sufficient parallel thickness ; from the centre d describe
the several concentric circles, which are the arrises of the moulding reauired.
To find the Sweep of the Moulding when the Plan %8 a Segment.
Fig. 2. Complete the semicircle ; then proceed as described under Fig. 1.
Figs. 3 and 4 show the method for bending a moulding round the inside, which is per-
formed the same as above.
The demonstration may eaaUy be conceived from the covering of a cone.
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APPENDIX.
WW^^K/V^W^VN^AMAM^AAA#^A ^^^ ^^*
STRENGTH OF TIMBER.
PROPOSITION I.
The Btrengths of the different pieces of timber, each of the same length and thiokneM,
are in proportion to the square of the deptli ; but if the tliickncHs nnd depth are both to be
considered, then the strength will be in proportion to the square of the depth, multiplied into
the thickness; and if all the three dimensions are taken jointly, then the weights that will
break each will be in proportion to the square of the depth multiplied into the thickneNS, and
divided by the length ; this is proved by the doctrine of mechanics. Hence a true rule will
appear for proportioning the strength of timbers to one another.
RULE.
Mvltijily the square of Vie depth of each piece of timber into tJie ViicJcneM ; and each product
being divided by the respective lengths^ will give live proportiouul strength of each.
EXAMPLE.
Suppose three pieces of timber, of the following dimensions :
The first, 6 inches deep, 3 inches thick, and 12 feet long.
The second, 5 inches deep, 4 inches thick, and 8 feet long.
The third, 9 inches deep, 8 inches thick, and 15 feet long. The comparative weight
that will break each piece is required. Ans. 9, 12i, and OH.
13 (97)
98 APPENDIX.
Tlierefore the weights that will break each are nearly in proportion to the numbers 9.
12, and 43, leaving out the fractions, in which you will observe, that the number 43 is almost
5 times the number 9 ; therefore the third piece of timber will ahnost bear 5 times as much
weight as the first , and the second piece nearly once and a third the weight of the first
piece; because the number 12 is once and a third greater than the number 9.
The timber is supposed to be everywhere of the same texture, otherwise these calcubi^
tions cannot hold true.
PROPOSITION II.
Given the length, breadth, and depth of a piece of timber ; to find the depth of another
piece whose length and breadth are given, so that it shall bear the same weight as the first
piece, or any number of times more.
RULE.
Multiply the square of the depth of the first piece into its hreadth, and divide that product hij its
length : multiply the quotient by the number of times as you would have the other piece to
carry more weight than the first, and multiply that by the length of the last pieceffatid dicidc
it by its width ; out of this last quotient extrojct the squ^are rooty which is tJie depth required,
EXAMPLE I.
Suppose a piece of timber 12 feet long, 6 inches deep, 4 inches thick; another piece 20
feet long, 5 inches thick ; requireth its depth, so that it shall bear twice the weight of the
first piece. Ans. 9*8 inches, nearly.
EXAMPLE IL
Suppose a piece of timber 14 feet long, 8 inches deep, 3 inches thick ; requireth the
depth of another piece 18 feet long, 4 inches thick, so that the last piece shall bear five times
as much weight as the first. Ans. 17*5 inches, nearly.
Note. — As the length of both pieces of timber is divisible by^the number 2, therefore half the length of each i
used insiead of the whole ; the answer will be the same*
PROPOSITION III.
Given the length, breadth, and depth, of a piece of timber; to find the breadth of ano-
thei piece whose length and depth is given, so that the last piece shall bear the same weight
as the first piece, or any number of times more.
APPENDIX. 99
RULE.
Midtiphj ili€ square of the depth of the first piece into its thicknees ; tJicU divided by its lengthy
vndtipljj tlie q',iotie7it by the number of times as tjoii would have the last piece bear more tfian
the first ; that being multiplied by the length of the last piece^ and divided by tfte square of
its depth J tliis quotient wUl be Hie breadth required.
EXAMPLE L
Given a piece of timber 12 feet long, 6 inches deep, 4 inches thick ; and another piece
16 feet long, 8 inches deep; requireth the thickness^ so that it shall bear twice as much
weight as the first piece. Arts. 6 inches.
EXAMPLE II.
Given a piece of timber 12 feet long, 5 inches deep, 3 inches thick ; and another piece
14 feet long, 6 inches deep ; requireth the thickness, so that the last piece may bear four
times as much weight as the first piece. Ans. 9*722 inches.
♦ PROPOSITION IV.
If the stress does not lie in the middle of the timl>er, but nearer to one end than the
other, the strength in the middle will be to the strength in any other part of the timber, as 1
divided by the square of half the length is to 1 divided by the rectangle of the two segments,
which are parted by the weight.
EXAMPLE I.
Suppose a piece of timber 20 feet long, the depth and width is immaterial ; suppose ttie
stress or weight to lie five feet distant from one of its ends, consequently from the other end
15 feet, then the above portion will be -^-r — tt^ ~ tt^r • r ^ i >- = ^- as the strength at fiv#»
^ 10 X 10 100 o^i'o lb ^
feet from the end is to the strength at the middle, or ten feet, or as ^-^r^ = 1 : -^- = 1 s-
° 100 10 3
Hence it appears that a piece of timber 20 feet long is one-third stronger at 5 feer
distance for the bearing, than it is in the middle, which is 10 feet, when cut in the above
proportion.
EXAMPLE IL
Suppose a piece of timber 30 feet long ; let the weight be applied 4 feet distant from one
end, or more properly from the place where it takes its bearing, then from the other end if
ATPENDIX.
be 26 feet, and the middle is 15 feet; then,
'226
1
4x26"
225
. 225_ _
225~ 104^^104
or nearly 2 ^.
169 169 , 169 ,.
169'I03'"'»''^106""'P'"P°''"°°
J stilted thus : 1 :
: 266: 428+, Jhs.
Hence it appears that a piece of timber 30 feet long will bear double the weight, and
one-sixth more, at four feet distance from one end, tlian it will do in the middle, which itt 15
feet distant.
EXAMPLE III.
Allowing that 266 pounds will break a beam 26 inches long, requireth the weight thai
will break the same beam when it lies at 5 inches from either end; then the distance to the
other end is 21 inches; 21 x 5 = 105, the half of 26 inches is 13.-. 13x 13 = 169; there-
lore the strength at the middle of the piece is to the strength at 5 inches from the end, as
1G9
105'
From this calculation it appears, that rather more than 428 pounds will break the beam
at 5 inches distance from one of its ends, if 266 pounds will break the same beam in the
middle.
Bj similar propositions the scantlings of any timber may be computed, so that they shall
sustain any given weight; for if the weight one piece will sustain be known, with its dimen-
sions, the weight that another piece will sustain, of any given dimensions, may also be
computed. The reader must observe, that although the foregoing rules are mathematicatly
true, yet it is impossible to account for knots, cross-grained wood, &c., such pieces being not
so strong as those which are straight i.i the grain ; and if care is not taken in choosing the
timber for a building, so that the grain of the timbers run nearly equal to one another, all
rules which can be laid down will Vie baffled, and consequently all rules for just proportion
will be useless in respect to its strength. It will be impossible, however, to estimate the
strength of timber fit for any building, or to have any true knowledge of its proportions,
without some rule; as without a rule everything must be done by mere conjecture.
Timber is much weakened by its own weight, except it stands perpendicular, which will
be shown in the following problems ; if a mortice is to be cut in the side of a piece of timber,
it will be much less weakened when cut near the top, than it will be if cut at the bottom,
provided the tenon is drove hard in to fill up the mortice.
The bending of timber will be nearly in proportion to the weight that is laid on it; no
beam ought to be trusted for any long time mth above one-third or one-fourth part of the
weight it will absolutely carry : for experiment proves, that a far less weight will break a i
piece of timber when hung to it for any considerable time, than what is sufiicient to break it
when first applied
APPENDIX. 101
PROBLEM I.
Saving the length and weight of a beam that can just support a given weight, to find'
ngth of another beam of the same scantling that shall just break with its own weight.
Let I = the length of the first beam,
L = the length of the second ;
a = the weight of the first beam,
w = the additional weight that will break it.
'1 because the weights that will break beams of the same scantling are reciprocally as
*ir lengths,
11 a Z
orefore — : — ::«? + -: Z = TF== the weight that will break the greater beam ; be-
l L 2 X
a
.ause u? + ^ is the whole weight that will break the lesser beam.
But the weights of beams of the same scantling are to one another as their lengths :
Whence, Z : X : : ^ : ^-j = Tfhalf the weight of the greater beam.
Now the beam cannot break by its own weight, unless the weight of the beam be equal
to the weight that will break it :
a
La to + 2 2w -{-a
Wherefore, = 1 = l.\L^a = 2wV-\-a Z^
2Z L 2L
.'. a : 2 ti? + a : : Z^ : i^, consequently y/ L^ = L = the length of the beam that can just
sustain its own weight.
PROBLEM XL
Having the weight of a beam that can just support a giveij weight in the middle, to find
the depth of another beam similar to the former, so that it shall just support its own weight.
Let d = the depth of the first beam ;
X = the depth of the second ;
a = the weight of the first beam ;
fv = the additional weight that will break the first beam ;
then will to-h^ov — ^ — == the whole weight that will break the lesser beam.
And because the weights that will break similar beams are as the squares of their lengtlis,
. /7« « 2ti?-f g 2a?^xt^-f g gg
102 APPENDIX.
the weights of similnr beams are as the cubes of their corresponding sides :
CL CI QC
Hence d^ : x^ : : ^ : ^-^ = W
ax^ 2x^w-\-x^a _ _ _
.'. 2^ = 2d^ .\ax = 2wd + ad
.'. a : a -{■ 2 w : : d : X = the depth required
As the weight of the lesser beam is to the weight of the lesser beam together with the
additional weight, so is the depth of the lesser beam to the depth of the greater beam.
Note. — Any other correepondiDg sides will answer the same purpose, for they are all proportioned to one another.
EXAMPLE.
Suppose a beam whose weight is one pound, and its length 10 feet, to carry a weight of
399-5 pounds, requireth the length of a beam similar to the former, of the same matter, so
that it shall break with its own weight.
here a = 1
and ti7 = 399-5
then a + 2 1/? = 800 = 1 + 2 X 3995
d = 10
Then by the last problem it will be 1 : 800 : 10 : 8000 = x for the length of a beam
that will break by its own weight.
PROBLEM III.
The weight and length of a piece of timber being given, and the additional weight that
will break it, to find the length of a piece of timber similar to the former, so that this last
piece of timber shall be the strongest possible :
Put I = the length of the piece given
w = half its weight,
W = the weight that will break it ;
X = the length required.
Then, because the weights that will break similar pieces of timber are in proportion to the
squares of their lengths,
j'.V :x^ : :W+ w : = ^ = the whole weight that breaks the beam ;
and because the weights of similar beams are as the cubes of their lengths, or any other
corresponding sides,
APPENDIX.
108
wx
then l^ : x^ : I to : -j^ the weight of the beam ;
consequently
I
Wx^-\-wx^^ wx
8
less -rj- is the weight that breaks the beam s= a maximum ;
l^ — l^
therefore its fluxion is nothing.
• that is, 2 Wx x + 2wxx ^ = nothing.
V
oTxr. o StVX ^. ^ ,^2W+2W
2 Tr+ 2tv = —rr— therefore, x = lx — ^
I ow
iience it ai)pears from the foregoing problems, that large timber is weakened in a much
greater proportion than small timber, even in similar pieces, therefore a proper allowance
must be made for the weight of the pieces, as I shall here show by an
EXAMPLE.
Suppose a beam 12 feet long, and a foot square, whose weight is 3 hundred weight, to be
capable of supporting 20 hundred weight, what weight will a beam 20 feet long, 15 inches
deep, and 12 thick, be able to support ?
12 inches square
12
144
12 '
15
15
12)1728
144
75
15
225
12
20)2700
135
But the weights of both beams are as their solid contents :
therefore 12 inches square
12
144
144 inches =: 12 feet long
576
576
144
20786 solid eootents of the 1st beam
15 deep
12 wide
180
240 length in incbei
7200
860
48200 solid contents of the 2d
104
20736: 43200 ::S
3
cwt.
20736)129600(6 ..
124416
lb.
28 =
= the weight
beam
of
API
the 2d
»ENDIX.
144 : : 135 : : 21-5 bj prop. 1.
21-5
67-5
135
270
5184
• « UXkj'T
112
12)2902-5
10368
5184
5184
12)241-875
20-15625
112
20736)580608(28
41472
31250
15625
15625
165888
165S88
17-50000
16
30
5
8-0
21 cwt. 56 lb. is the weight that will break the first beam, and 20 cwt. 17 lb. 8 oz. the
weight that will break the second beam ; deduct out of these half their own weight.
20::17::8
3::14::0 half
1 / . . . o . . o
Now 20 cwt. is the additional weight that will break the first beam; and 17 cwt. 3 lb.
8 oz. the weight that will break the second : in which the reader will observe, that 10 : : 3 : : 8
has a much less proportion to 20, than 20 cwt. 17 lb. 8 oz. has to 21 : : 56. From these
examples, the reader may see that a proper allowance ought to be made for all horizontal
beams ; that is, half the weights of beams ought to be deducted out of the whole weight that
they will carry, and that will give the weight that each piece will bear.
If several pieces of timber of the same scantling and length are applied one above ano-
ther, and supported by props at each end, they will be no stronger than if they were laid
side by side ; or this, which is the same thing, the pieces that are applied one above another
are no stronger than one single piepe whose width is the width of the several pieces collected
into one, and its depth the depth of one of the pieces ; it is therefore useless to cut a piece
of timber lengthways, and apply the pieces so cut one above another, for these pieces are not
so strong as before, even if bolted.
EXAMPLE.
Suppose a girder 16 inches deep, 12 inches thick, the length is immaterial, and let the
depth be cut lengthways in two equal pieces; then will each piece be 8 inches deep, and 12
I.
APPENDIX. 106
inches thick. Now, according to the rule of proportioning timber, the square of 16 inches,
that is, the depth before it was cut, is 256, and the square of 8 Inches is 64 ; but twice 64 is
only 128, therefore it appears that the two pieces applied one above another, are but half the
strength of the solid piece, because 256 is double 128.
If a girder be cut lengthways in a perpendicular direction, the ends turned contrary, and
then bolted together, it will be but very little stronger than before it was cut ; for although
the ends being turned give to the girder an equal strength throughout, yet wherever a bolt
is, there it will be weaker, and it is very doubtful whether the girder will be any stronger for
this process of sawing and bolting ; and I sa^^ this from experience.
•
If there are two pieces of timber of an equal scantling (PI. 51, Fig. -B), the one lying
horizontal, and the other inclined, the horizontal piece being supported at the points e and /,
and the inclined piece at c and rf, perpendicularly over e and/, according to the principles of
mechanics, these pieces will be equally strong. But, to reason a little on this matter, let it
be considered, that although the inclined piece D is longer, yet the weight has less effect upon
It when placed in the middle, than the weight at h has upon the horizontal piece C7, the
weights being the same ; it is therefore reasonable to conclude, that in these positions the one
will bear equal to the other.
The foregoing rules will be found of excellent use when timber is wanted to support a
great weight ; for, by knowing the superincumbent weight, the strength may be computed to
a great degree of exactness, so that it shall be able to support the weight required. The
consequence is as bad when there is too much timber, as when there is too little, for nothing
is more requisite than a just proportion throughout the whole building, so that the strength
of every part shall always be in proportion to the stress ; for when there is more strength
given to some pieces than others, it encumbers the building, and consequently the foundations
are less capable of supporting the superstructure.
No judicious person, who has the care of constructing buildings, should rely on tables
of scantlings, such as are commonly in books; for example, in story posts the scantlings,
according to several authors, are as follows :
For 9 feet high 6 inches square.
12 8
15 10
18 12
14
lOG APPENDIX.
Now, according to this table, the scnntlinga nre increased in proportion to the height;
but there ia no propriety in this, for each of these will bear weight iu proportion to the
number 9, 16, 25, and 36, that ia, iij proportion to the square of their heights, 36 being 4
times 9 ; therefore the piece that is 18 feet long, will bear ibur times as much weight as ihat
piece which is 9 feet long; but the 9 feet piece may have a much greater weight to carry
than an 18 feet piece, suppose double: in this case it umst be near 12 inches square instead
of 6. The same ia also to be observed in breast-summers, and in floors where they are wanted
to support a great weight; but in common buildings, where there are only customary weighta
to support, the common tables for floors will be near enough for practice.
To conclude the subject, it may be proper to notice the following observations which
several authors have judiciously made, viz. ; that in all timber there is moisture, wherefore
all bearing timber ought to have a moderate camber, or roundness on the upper side, for till
that moisture is dried out, the timber will swag with its own weight.
But then observe, that it is best to truss girders when they are fresh sawn out, for by
their drying and shrinking, the trusses become more and more tight.
That all beams or ties be cut, or iu framing forced to a roundness, such as an inch in
twenty feet iu length, and that principal rafters also be cut or forced in framing, as before;
because all joists, though ever so well framed, by the shrinking of the timber and weight of
til'- covering will swng, somelimes so much as not only to be visible, but to offend the eye:
by lliis precaution the tru-^H will always appear well.
Likewise observe, that all ciise-bays, either in floors or roofs, do not exceed twelve feet
if possible ; that is, do not let your joiutri in floors exceed twelve feet, nor your purliues in
roofs, &c., but rather let their liearing be eight, nine, or ten feet. This should be regarded in
ibrmiiig the plan.
Also, in bridging floors, do not place your binding or strong joists above three, four, or
five feet apart, and take care that your bridging or common joists are not above ten or twelve
inches apart, that is, between one joist and another.
Also, in fitting down lie-beams upon the wall plates, never make your cocking too large,
nor yet too near the outside of the wall plate, for the grain of the wood being cut across in
the tie-beam, the piece that remains upon its end will be apt to split off, but keeping it near
the inside will tend to secure it.
Likewise observe, never to make double tenons for bearing uses, such as binding joists.
APPENDIX. 107
st.s, or purlines; for, in the first place, it very much weakens whatever you frame
iin'l in the second place, it is a rarity to have a dniught to both tenons, that is, to
iih joints close j for the pin in passing through both tenonsj if there is a draught in
bend so much, that unless it he as tough bb wire, it must needs break in driving,
u^'queiitly do more hurt than good.
A ill be much stronger if the purlines are notched above the principal rafters, than
' ' IVamed into the side of the principals; for by this means, when any weight is
^1^ in the middle of the purline, it cannot bend, being confined by the other rafters; and
do, the sides of the other raflers must needs bend along with it; consequently it has the
i^gth of all the other rafters sideways added to it.
^^M
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